Write in Standard Form y^2-4y+12x+4=0

$y^{2} - 4 y + 12 x + 4 = 0$

Solve for $y$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 4$ , and $c = 12 x + 4$ into the quadratic formula and solve for $y$ .

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

Simplify.

Simplify the numerator.

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (12 x + 4)$ .

Factor $- 4$ out of ${(- 4)}^{2}$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 \cdot 1 \cdot (12 x + 4)}}{2 \cdot 1}$

Factor $- 4$ out of $- 4 \cdot 1 \cdot (12 x + 4)$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 (1 \cdot (12 x + 4))}}{2 \cdot 1}$

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (12 x + 4))$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 1 \cdot (12 x + 4))}}{2 \cdot 1}$

Multiply $12 x + 4$ by $1$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 12 x + 4)}}{2 \cdot 1}$

Add $- 4$ and $4$ .

$y = \frac{4 \pm \sqrt{- 4 (12 x + 0)}}{2 \cdot 1}$

Add $12 x$ and $0$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot (12 x)}}{2 \cdot 1}$

Multiply $- 4$ by $12$ .

$y = \frac{4 \pm \sqrt{- 48 x}}{2 \cdot 1}$

Rewrite $- 48 x$ as ${(2^{2})}^{2} \cdot (- 3 x)$ .

Factor $16$ out of $- 48$ .

$y = \frac{4 \pm \sqrt{16 (- 3) x}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

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

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

Add parentheses.

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

Pull terms out from under the radical.

$y = \frac{4 \pm 2^{2} \sqrt{- 3 x}}{2 \cdot 1}$

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

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

Multiply $2$ by $1$ .

$y = \frac{4 \pm 4 \sqrt{- 3 x}}{2}$

Simplify $\frac{4 \pm 4 \sqrt{- 3 x}}{2}$ .

$y = 2 \pm 2 \sqrt{- 3 x}$

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

Simplify the numerator.

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (12 x + 4)$ .

Factor $- 4$ out of ${(- 4)}^{2}$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 \cdot 1 \cdot (12 x + 4)}}{2 \cdot 1}$

Factor $- 4$ out of $- 4 \cdot 1 \cdot (12 x + 4)$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 (1 \cdot (12 x + 4))}}{2 \cdot 1}$

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (12 x + 4))$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 1 \cdot (12 x + 4))}}{2 \cdot 1}$

Multiply $12 x + 4$ by $1$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 12 x + 4)}}{2 \cdot 1}$

Add $- 4$ and $4$ .

$y = \frac{4 \pm \sqrt{- 4 (12 x + 0)}}{2 \cdot 1}$

Add $12 x$ and $0$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot (12 x)}}{2 \cdot 1}$

Multiply $- 4$ by $12$ .

$y = \frac{4 \pm \sqrt{- 48 x}}{2 \cdot 1}$

Rewrite $- 48 x$ as ${(2^{2})}^{2} \cdot (- 3 x)$ .

Factor $16$ out of $- 48$ .

$y = \frac{4 \pm \sqrt{16 (- 3) x}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

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

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

Add parentheses.

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

Pull terms out from under the radical.

$y = \frac{4 \pm 2^{2} \sqrt{- 3 x}}{2 \cdot 1}$

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

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

Multiply $2$ by $1$ .

$y = \frac{4 \pm 4 \sqrt{- 3 x}}{2}$

Simplify $\frac{4 \pm 4 \sqrt{- 3 x}}{2}$ .

$y = 2 \pm 2 \sqrt{- 3 x}$

Change the $\pm$ to $+$ .

$y = 2 + 2 \sqrt{- 3 x}$

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

Simplify the numerator.

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (12 x + 4)$ .

Factor $- 4$ out of ${(- 4)}^{2}$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 \cdot 1 \cdot (12 x + 4)}}{2 \cdot 1}$

Factor $- 4$ out of $- 4 \cdot 1 \cdot (12 x + 4)$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 (1 \cdot (12 x + 4))}}{2 \cdot 1}$

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (12 x + 4))$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 1 \cdot (12 x + 4))}}{2 \cdot 1}$

Multiply $12 x + 4$ by $1$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 12 x + 4)}}{2 \cdot 1}$

Add $- 4$ and $4$ .

$y = \frac{4 \pm \sqrt{- 4 (12 x + 0)}}{2 \cdot 1}$

Add $12 x$ and $0$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot (12 x)}}{2 \cdot 1}$

Multiply $- 4$ by $12$ .

$y = \frac{4 \pm \sqrt{- 48 x}}{2 \cdot 1}$

Rewrite $- 48 x$ as ${(2^{2})}^{2} \cdot (- 3 x)$ .

Factor $16$ out of $- 48$ .

$y = \frac{4 \pm \sqrt{16 (- 3) x}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

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

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

Add parentheses.

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

Pull terms out from under the radical.

$y = \frac{4 \pm 2^{2} \sqrt{- 3 x}}{2 \cdot 1}$

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

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

Multiply $2$ by $1$ .

$y = \frac{4 \pm 4 \sqrt{- 3 x}}{2}$

Simplify $\frac{4 \pm 4 \sqrt{- 3 x}}{2}$ .

$y = 2 \pm 2 \sqrt{- 3 x}$

Change the $\pm$ to $-$ .

$y = 2 - 2 \sqrt{- 3 x}$

The final answer is the combination of both solutions.

$y = 2 + 2 \sqrt{- 3 x}$

$y = 2 - 2 \sqrt{- 3 x}$

$y = 2 + 2 \sqrt{- 3 x}$

$y = 2 - 2 \sqrt{- 3 x}$

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$y = a x^{2} + b x + c$

The standard form is $2 + 2 \sqrt{- 3 x}, 2 - 2 \sqrt{- 3 x}$ .

$y = 2 + 2 \sqrt{- 3 x}$

$y = 2 - 2 \sqrt{- 3 x}$

Write in Standard Form y^2-4y+12x+4=0

Related Questions

Support LCMGCF.com by disabling your adblocker.