Evaluate xy(3x+2y)=4

$x y (3 x + 2 y) = 4$

Simplify $x y (3 x + 2 y)$ .

Simplify by multiplying through.

Apply the distributive property.

$x y (3 x) + x y (2 y) = 4$

Reorder.

Rewrite using the commutative property of multiplication.

$3 x y x + x y (2 y) = 4$

Rewrite using the commutative property of multiplication.

$3 x y x + 2 x y \cdot y = 4$

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$3 (x \cdot x) y + 2 x y \cdot y = 4$

Multiply $x$ by $x$ .

$3 x^{2} y + 2 x y \cdot y = 4$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$3 x^{2} y + 2 x (y \cdot y) = 4$

Multiply $y$ by $y$ .

$3 x^{2} y + 2 x y^{2} = 4$

Subtract $4$ from both sides of the equation.

$3 x^{2} y + 2 x y^{2} - 4 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 3 y$ , $b = 2 y^{2}$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{- 2 y^{2} \pm \sqrt{{(2 y^{2})}^{2} - 4 \cdot (3 y \cdot - 4)}}{2 (3 y)}$

Simplify.

Simplify the numerator.

Add parentheses.

$x = \frac{- 2 y^{2} \pm \sqrt{{(2 y^{2})}^{2} - 4 \cdot (3 (y \cdot - 4))}}{2 \cdot (3 y)}$

Let $u = 3 (y \cdot - 4)$ . Substitute $u$ for all occurrences of $3 (y \cdot - 4)$ .

Apply the product rule to $2 y^{2}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} {(y^{2})}^{2} - 4 \cdot u}}{2 \cdot (3 y)}$

Raise $2$ to the power of $2$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 {(y^{2})}^{2} - 4 \cdot u}}{2 \cdot (3 y)}$

Multiply the exponents in ${(y^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{2 \cdot 2} - 4 \cdot u}}{2 \cdot (3 y)}$

Multiply $2$ by $2$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{4} - 4 \cdot u}}{2 \cdot (3 y)}$

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{4} - 4 u}}{2 \cdot (3 y)}$

Factor $4$ out of $4 y^{4} - 4 u$ .

Factor $4$ out of $4 y^{4}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4}) - 4 u}}{2 \cdot (3 y)}$

Factor $4$ out of $- 4 u$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4}) + 4 (- u)}}{2 \cdot (3 y)}$

Factor $4$ out of $4 (y^{4}) + 4 (- u)$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - u)}}{2 \cdot (3 y)}$

Replace all occurrences of $u$ with $3 (y \cdot - 4)$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (3 (y \cdot - 4)))}}{2 \cdot (3 y)}$

Simplify each term.

Move $- 4$ to the left of $y$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (3 (- 4 \cdot y)))}}{2 \cdot (3 y)}$

Multiply $- 4$ by $3$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (- 12 y))}}{2 \cdot (3 y)}$

Multiply $- 12$ by $- 1$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} + 12 y)}}{2 \cdot (3 y)}$

Factor $y$ out of $y^{4} + 12 y$ .

Factor $y$ out of $y^{4}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y \cdot y^{3} + 12 y)}}{2 \cdot (3 y)}$

Factor $y$ out of $12 y$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y \cdot y^{3} + y \cdot 12)}}{2 \cdot (3 y)}$

Factor $y$ out of $y \cdot y^{3} + y \cdot 12$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y (y^{3} + 12))}}{2 \cdot (3 y)}$

$x = \frac{- 2 y^{2} \pm \sqrt{4 y (y^{3} + 12)}}{2 \cdot (3 y)}$

Rewrite $4 y (y^{3} + 12)$ as $2^{2} (y (y^{3} + 12))$ .

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} y (y^{3} + 12)}}{2 \cdot (3 y)}$

Add parentheses.

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} (y (y^{3} + 12))}}{2 \cdot (3 y)}$

Pull terms out from under the radical.

$x = \frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{2 \cdot (3 y)}$

Multiply $2$ by $3$ .

$x = \frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{6 y}$

Simplify $\frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{6 y}$ .

$x = \frac{- y^{2} \pm \sqrt{y (y^{3} + 12)}}{3 y}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Add parentheses.

$x = \frac{- 2 y^{2} \pm \sqrt{{(2 y^{2})}^{2} - 4 \cdot (3 (y \cdot - 4))}}{2 \cdot (3 y)}$

Let $u = 3 (y \cdot - 4)$ . Substitute $u$ for all occurrences of $3 (y \cdot - 4)$ .

Apply the product rule to $2 y^{2}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} {(y^{2})}^{2} - 4 \cdot u}}{2 \cdot (3 y)}$

Raise $2$ to the power of $2$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 {(y^{2})}^{2} - 4 \cdot u}}{2 \cdot (3 y)}$

Multiply the exponents in ${(y^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{2 \cdot 2} - 4 \cdot u}}{2 \cdot (3 y)}$

Multiply $2$ by $2$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{4} - 4 \cdot u}}{2 \cdot (3 y)}$

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{4} - 4 u}}{2 \cdot (3 y)}$

Factor $4$ out of $4 y^{4} - 4 u$ .

Factor $4$ out of $4 y^{4}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4}) - 4 u}}{2 \cdot (3 y)}$

Factor $4$ out of $- 4 u$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4}) + 4 (- u)}}{2 \cdot (3 y)}$

Factor $4$ out of $4 (y^{4}) + 4 (- u)$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - u)}}{2 \cdot (3 y)}$

Replace all occurrences of $u$ with $3 (y \cdot - 4)$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (3 (y \cdot - 4)))}}{2 \cdot (3 y)}$

Simplify each term.

Move $- 4$ to the left of $y$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (3 (- 4 \cdot y)))}}{2 \cdot (3 y)}$

Multiply $- 4$ by $3$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (- 12 y))}}{2 \cdot (3 y)}$

Multiply $- 12$ by $- 1$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} + 12 y)}}{2 \cdot (3 y)}$

Factor $y$ out of $y^{4} + 12 y$ .

Factor $y$ out of $y^{4}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y \cdot y^{3} + 12 y)}}{2 \cdot (3 y)}$

Factor $y$ out of $12 y$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y \cdot y^{3} + y \cdot 12)}}{2 \cdot (3 y)}$

Factor $y$ out of $y \cdot y^{3} + y \cdot 12$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y (y^{3} + 12))}}{2 \cdot (3 y)}$

$x = \frac{- 2 y^{2} \pm \sqrt{4 y (y^{3} + 12)}}{2 \cdot (3 y)}$

Rewrite $4 y (y^{3} + 12)$ as $2^{2} (y (y^{3} + 12))$ .

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} y (y^{3} + 12)}}{2 \cdot (3 y)}$

Add parentheses.

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} (y (y^{3} + 12))}}{2 \cdot (3 y)}$

Pull terms out from under the radical.

$x = \frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{2 \cdot (3 y)}$

Multiply $2$ by $3$ .

$x = \frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{6 y}$

Simplify $\frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{6 y}$ .

$x = \frac{- y^{2} \pm \sqrt{y (y^{3} + 12)}}{3 y}$

Change the $\pm$ to $+$ .

$x = \frac{- y^{2} + \sqrt{y (y^{3} + 12)}}{3 y}$

Factor $- 1$ out of $- y^{2}$ .

$x = \frac{- (y^{2}) + \sqrt{y (y^{3} + 12)}}{3 y}$

Factor $- 1$ out of $\sqrt{y (y^{3} + 12)}$ .

$x = \frac{- (y^{2}) - 1 (- \sqrt{y (y^{3} + 12)})}{3 y}$

Factor $- 1$ out of $- (y^{2}) - 1 (- \sqrt{y (y^{3} + 12)})$ .

$x = \frac{- (y^{2} - \sqrt{y (y^{3} + 12)})}{3 y}$

Rewrite $- (y^{2} - \sqrt{y (y^{3} + 12)})$ as $- 1 (y^{2} - \sqrt{y (y^{3} + 12)})$ .

$x = \frac{- 1 (y^{2} - \sqrt{y (y^{3} + 12)})}{3 y}$

Move the negative in front of the fraction.

$x = - \frac{y^{2} - \sqrt{y (y^{3} + 12)}}{3 y}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Add parentheses.

$x = \frac{- 2 y^{2} \pm \sqrt{{(2 y^{2})}^{2} - 4 \cdot (3 (y \cdot - 4))}}{2 \cdot (3 y)}$

Let $u = 3 (y \cdot - 4)$ . Substitute $u$ for all occurrences of $3 (y \cdot - 4)$ .

Apply the product rule to $2 y^{2}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} {(y^{2})}^{2} - 4 \cdot u}}{2 \cdot (3 y)}$

Raise $2$ to the power of $2$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 {(y^{2})}^{2} - 4 \cdot u}}{2 \cdot (3 y)}$

Multiply the exponents in ${(y^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{2 \cdot 2} - 4 \cdot u}}{2 \cdot (3 y)}$

Multiply $2$ by $2$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{4} - 4 \cdot u}}{2 \cdot (3 y)}$

$x = \frac{- 2 y^{2} \pm \sqrt{4 y^{4} - 4 u}}{2 \cdot (3 y)}$

Factor $4$ out of $4 y^{4} - 4 u$ .

Factor $4$ out of $4 y^{4}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4}) - 4 u}}{2 \cdot (3 y)}$

Factor $4$ out of $- 4 u$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4}) + 4 (- u)}}{2 \cdot (3 y)}$

Factor $4$ out of $4 (y^{4}) + 4 (- u)$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - u)}}{2 \cdot (3 y)}$

Replace all occurrences of $u$ with $3 (y \cdot - 4)$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (3 (y \cdot - 4)))}}{2 \cdot (3 y)}$

Simplify each term.

Move $- 4$ to the left of $y$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (3 (- 4 \cdot y)))}}{2 \cdot (3 y)}$

Multiply $- 4$ by $3$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} - (- 12 y))}}{2 \cdot (3 y)}$

Multiply $- 12$ by $- 1$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y^{4} + 12 y)}}{2 \cdot (3 y)}$

Factor $y$ out of $y^{4} + 12 y$ .

Factor $y$ out of $y^{4}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y \cdot y^{3} + 12 y)}}{2 \cdot (3 y)}$

Factor $y$ out of $12 y$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y \cdot y^{3} + y \cdot 12)}}{2 \cdot (3 y)}$

Factor $y$ out of $y \cdot y^{3} + y \cdot 12$ .

$x = \frac{- 2 y^{2} \pm \sqrt{4 (y (y^{3} + 12))}}{2 \cdot (3 y)}$

$x = \frac{- 2 y^{2} \pm \sqrt{4 y (y^{3} + 12)}}{2 \cdot (3 y)}$

Rewrite $4 y (y^{3} + 12)$ as $2^{2} (y (y^{3} + 12))$ .

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} y (y^{3} + 12)}}{2 \cdot (3 y)}$

Add parentheses.

$x = \frac{- 2 y^{2} \pm \sqrt{2^{2} (y (y^{3} + 12))}}{2 \cdot (3 y)}$

Pull terms out from under the radical.

$x = \frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{2 \cdot (3 y)}$

Multiply $2$ by $3$ .

$x = \frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{6 y}$

Simplify $\frac{- 2 y^{2} \pm 2 \sqrt{y (y^{3} + 12)}}{6 y}$ .

$x = \frac{- y^{2} \pm \sqrt{y (y^{3} + 12)}}{3 y}$

Change the $\pm$ to $-$ .

$x = \frac{- y^{2} - \sqrt{y (y^{3} + 12)}}{3 y}$

Factor $- 1$ out of $- y^{2}$ .

$x = \frac{- (y^{2}) - \sqrt{y (y^{3} + 12)}}{3 y}$

Factor $- 1$ out of $- \sqrt{y (y^{3} + 12)}$ .

$x = \frac{- (y^{2}) - (\sqrt{y (y^{3} + 12)})}{3 y}$

Factor $- 1$ out of $- (y^{2}) - (\sqrt{y (y^{3} + 12)})$ .

$x = \frac{- (y^{2} + \sqrt{y (y^{3} + 12)})}{3 y}$

Rewrite $- (y^{2} + \sqrt{y (y^{3} + 12)})$ as $- 1 (y^{2} + \sqrt{y (y^{3} + 12)})$ .

$x = \frac{- 1 (y^{2} + \sqrt{y (y^{3} + 12)})}{3 y}$

Move the negative in front of the fraction.

$x = - \frac{y^{2} + \sqrt{y (y^{3} + 12)}}{3 y}$

The final answer is the combination of both solutions.

$x = - \frac{y^{2} - \sqrt{y (y^{3} + 12)}}{3 y}$

$x = - \frac{y^{2} + \sqrt{y (y^{3} + 12)}}{3 y}$

Evaluate xy(3x+2y)=4

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