Solve for x 2x(x-4)-3(x+5)=x(1-x)-16

$2 x (x - 4) - 3 (x + 5) = x (1 - x) - 16$

Step 1

Simplify $2 x (x - 4) - 3 (x + 5)$ .

Simplify each term.

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 4 - 3 (x + 5) = x (1 - x) - 16$

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

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 4 - 3 (x + 5) = x (1 - x) - 16$

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 4 - 3 (x + 5) = x (1 - x) - 16$

Multiply $- 4$ by $2$ .

$2 x^{2} - 8 x - 3 (x + 5) = x (1 - x) - 16$

Apply the distributive property.

$2 x^{2} - 8 x - 3 x - 3 \cdot 5 = x (1 - x) - 16$

Multiply $- 3$ by $5$ .

$2 x^{2} - 8 x - 3 x - 15 = x (1 - x) - 16$

Subtract $3 x$ from $- 8 x$ .

$2 x^{2} - 11 x - 15 = x (1 - x) - 16$

Step 2

Simplify each term.

Apply the distributive property.

$2 x^{2} - 11 x - 15 = x \cdot 1 + x (- x) - 16$

Multiply $x$ by $1$ .

$2 x^{2} - 11 x - 15 = x + x (- x) - 16$

Rewrite using the commutative property of multiplication.

$2 x^{2} - 11 x - 15 = x - x \cdot x - 16$

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

Move $x$ .

$2 x^{2} - 11 x - 15 = x - (x \cdot x) - 16$

Multiply $x$ by $x$ .

$2 x^{2} - 11 x - 15 = x - x^{2} - 16$

Step 3

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$2 x^{2} - 11 x - 15 - x = - x^{2} - 16$

Add $x^{2}$ to both sides of the equation.

$2 x^{2} - 11 x - 15 - x + x^{2} = - 16$

Add $2 x^{2}$ and $x^{2}$ .

$3 x^{2} - 11 x - 15 - x = - 16$

Subtract $x$ from $- 11 x$ .

$3 x^{2} - 12 x - 15 = - 16$

Step 4

Move all terms to the left side of the equation and simplify.

Add $16$ to both sides of the equation.

$3 x^{2} - 12 x - 15 + 16 = 0$

Add $- 15$ and $16$ .

$3 x^{2} - 12 x + 1 = 0$

Step 5

Use the quadratic formula to find the solutions.

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

Step 6

Substitute the values $a = 3$ , $b = - 12$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (3 \cdot 1)}}{2 \cdot 3}$

Step 7

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 1$ .

Multiply $- 4$ by $3$ .

$x = \frac{12 \pm \sqrt{144 - 12 \cdot 1}}{2 \cdot 3}$

Multiply $- 12$ by $1$ .

$x = \frac{12 \pm \sqrt{144 - 12}}{2 \cdot 3}$

Subtract $12$ from $144$ .

$x = \frac{12 \pm \sqrt{132}}{2 \cdot 3}$

Rewrite $132$ as $2^{2} \cdot 33$ .

Factor $4$ out of $132$ .

$x = \frac{12 \pm \sqrt{4 (33)}}{2 \cdot 3}$

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

$x = \frac{12 \pm \sqrt{2^{2} \cdot 33}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{12 \pm 2 \sqrt{33}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{12 \pm 2 \sqrt{33}}{6}$

Simplify $\frac{12 \pm 2 \sqrt{33}}{6}$ .

$x = \frac{6 \pm \sqrt{33}}{3}$

Step 8

The final answer is the combination of both solutions.

$x = \frac{6 + \sqrt{33}}{3}, \frac{6 - \sqrt{33}}{3}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$x = \frac{6 + \sqrt{33}}{3}, \frac{6 - \sqrt{33}}{3}$

Decimal Form:

$x = 3.91485421 \dots, 0.08514578 \dots$

Solve for x 2x(x-4)-3(x+5)=x(1-x)-16

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