Write in Standard Form 4x-y^2-2y-33=0

$4 x - y^{2} - 2 y - 33 = 0$

Step 1

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 $a = - 1$ , $b = - 2$ , and $c = 4 x - 33$ into the quadratic formula and solve for $y$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (- 1 \cdot (4 x - 33))}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot - 1 \cdot (4 x - 33)}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$y = \frac{2 \pm \sqrt{4 + 4 \cdot (4 x - 33)}}{2 \cdot - 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 + 4 (4 x) + 4 \cdot - 33}}{2 \cdot - 1}$

Multiply $4$ by $4$ .

$y = \frac{2 \pm \sqrt{4 + 16 x + 4 \cdot - 33}}{2 \cdot - 1}$

Multiply $4$ by $- 33$ .

$y = \frac{2 \pm \sqrt{4 + 16 x - 132}}{2 \cdot - 1}$

Subtract $132$ from $4$ .

$y = \frac{2 \pm \sqrt{16 x - 128}}{2 \cdot - 1}$

Factor $16$ out of $16 x - 128$ .

Factor $16$ out of $16 x$ .

$y = \frac{2 \pm \sqrt{16 (x) - 128}}{2 \cdot - 1}$

Factor $16$ out of $- 128$ .

$y = \frac{2 \pm \sqrt{16 x + 16 \cdot - 8}}{2 \cdot - 1}$

Factor $16$ out of $16 x + 16 \cdot - 8$ .

$y = \frac{2 \pm \sqrt{16 (x - 8)}}{2 \cdot - 1}$

Rewrite $16 (x - 8)$ as ${(2^{2})}^{2} (x - 8)$ .

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

$y = \frac{2 \pm \sqrt{4^{2} (x - 8)}}{2 \cdot - 1}$

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

$y = \frac{2 \pm \sqrt{{(2^{2})}^{2} (x - 8)}}{2 \cdot - 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2^{2} \sqrt{x - 8}}{2 \cdot - 1}$

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

$y = \frac{2 \pm 4 \sqrt{x - 8}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$y = \frac{2 \pm 4 \sqrt{x - 8}}{- 2}$

Simplify $\frac{2 \pm 4 \sqrt{x - 8}}{- 2}$ .

$y = \frac{1 \pm 2 \sqrt{x - 8}}{- 1}$

Move the negative one from the denominator of $\frac{1 \pm 2 \sqrt{x - 8}}{- 1}$ .

$y = - 1 \cdot (1 \pm 2 \sqrt{x - 8})$

Rewrite $- 1 \cdot (1 \pm 2 \sqrt{x - 8})$ as $- (1 \pm 2 \sqrt{x - 8})$ .

$y = - (1 \pm 2 \sqrt{x - 8})$

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

Simplify the numerator.

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot - 1 \cdot (4 x - 33)}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$y = \frac{2 \pm \sqrt{4 + 4 \cdot (4 x - 33)}}{2 \cdot - 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 + 4 (4 x) + 4 \cdot - 33}}{2 \cdot - 1}$

Multiply $4$ by $4$ .

$y = \frac{2 \pm \sqrt{4 + 16 x + 4 \cdot - 33}}{2 \cdot - 1}$

Multiply $4$ by $- 33$ .

$y = \frac{2 \pm \sqrt{4 + 16 x - 132}}{2 \cdot - 1}$

Subtract $132$ from $4$ .

$y = \frac{2 \pm \sqrt{16 x - 128}}{2 \cdot - 1}$

Factor $16$ out of $16 x - 128$ .

Factor $16$ out of $16 x$ .

$y = \frac{2 \pm \sqrt{16 (x) - 128}}{2 \cdot - 1}$

Factor $16$ out of $- 128$ .

$y = \frac{2 \pm \sqrt{16 x + 16 \cdot - 8}}{2 \cdot - 1}$

Factor $16$ out of $16 x + 16 \cdot - 8$ .

$y = \frac{2 \pm \sqrt{16 (x - 8)}}{2 \cdot - 1}$

Rewrite $16 (x - 8)$ as ${(2^{2})}^{2} (x - 8)$ .

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

$y = \frac{2 \pm \sqrt{4^{2} (x - 8)}}{2 \cdot - 1}$

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

$y = \frac{2 \pm \sqrt{{(2^{2})}^{2} (x - 8)}}{2 \cdot - 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2^{2} \sqrt{x - 8}}{2 \cdot - 1}$

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

$y = \frac{2 \pm 4 \sqrt{x - 8}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$y = \frac{2 \pm 4 \sqrt{x - 8}}{- 2}$

Simplify $\frac{2 \pm 4 \sqrt{x - 8}}{- 2}$ .

$y = \frac{1 \pm 2 \sqrt{x - 8}}{- 1}$

Move the negative one from the denominator of $\frac{1 \pm 2 \sqrt{x - 8}}{- 1}$ .

$y = - 1 \cdot (1 \pm 2 \sqrt{x - 8})$

Rewrite $- 1 \cdot (1 \pm 2 \sqrt{x - 8})$ as $- (1 \pm 2 \sqrt{x - 8})$ .

$y = - (1 \pm 2 \sqrt{x - 8})$

Change the $\pm$ to $+$ .

$y = - (1 + 2 \sqrt{x - 8})$

Apply the distributive property.

$y = - 1 \cdot 1 - (2 \sqrt{x - 8})$

Multiply $- 1$ by $1$ .

$y = - 1 - (2 \sqrt{x - 8})$

Multiply $2$ by $- 1$ .

$y = - 1 - 2 \sqrt{x - 8}$

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

Simplify the numerator.

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot - 1 \cdot (4 x - 33)}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$y = \frac{2 \pm \sqrt{4 + 4 \cdot (4 x - 33)}}{2 \cdot - 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 + 4 (4 x) + 4 \cdot - 33}}{2 \cdot - 1}$

Multiply $4$ by $4$ .

$y = \frac{2 \pm \sqrt{4 + 16 x + 4 \cdot - 33}}{2 \cdot - 1}$

Multiply $4$ by $- 33$ .

$y = \frac{2 \pm \sqrt{4 + 16 x - 132}}{2 \cdot - 1}$

Subtract $132$ from $4$ .

$y = \frac{2 \pm \sqrt{16 x - 128}}{2 \cdot - 1}$

Factor $16$ out of $16 x - 128$ .

Factor $16$ out of $16 x$ .

$y = \frac{2 \pm \sqrt{16 (x) - 128}}{2 \cdot - 1}$

Factor $16$ out of $- 128$ .

$y = \frac{2 \pm \sqrt{16 x + 16 \cdot - 8}}{2 \cdot - 1}$

Factor $16$ out of $16 x + 16 \cdot - 8$ .

$y = \frac{2 \pm \sqrt{16 (x - 8)}}{2 \cdot - 1}$

Rewrite $16 (x - 8)$ as ${(2^{2})}^{2} (x - 8)$ .

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

$y = \frac{2 \pm \sqrt{4^{2} (x - 8)}}{2 \cdot - 1}$

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

$y = \frac{2 \pm \sqrt{{(2^{2})}^{2} (x - 8)}}{2 \cdot - 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2^{2} \sqrt{x - 8}}{2 \cdot - 1}$

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

$y = \frac{2 \pm 4 \sqrt{x - 8}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$y = \frac{2 \pm 4 \sqrt{x - 8}}{- 2}$

Simplify $\frac{2 \pm 4 \sqrt{x - 8}}{- 2}$ .

$y = \frac{1 \pm 2 \sqrt{x - 8}}{- 1}$

Move the negative one from the denominator of $\frac{1 \pm 2 \sqrt{x - 8}}{- 1}$ .

$y = - 1 \cdot (1 \pm 2 \sqrt{x - 8})$

Rewrite $- 1 \cdot (1 \pm 2 \sqrt{x - 8})$ as $- (1 \pm 2 \sqrt{x - 8})$ .

$y = - (1 \pm 2 \sqrt{x - 8})$

Change the $\pm$ to $-$ .

$y = - (1 - 2 \sqrt{x - 8})$

Apply the distributive property.

$y = - 1 \cdot 1 - (- 2 \sqrt{x - 8})$

Multiply $- 1$ by $1$ .

$y = - 1 - (- 2 \sqrt{x - 8})$

Multiply $- 2$ by $- 1$ .

$y = - 1 + 2 \sqrt{x - 8}$

The final answer is the combination of both solutions.

$y = - 1 - 2 \sqrt{x - 8}$

$y = - 1 + 2 \sqrt{x - 8}$

$y = - 1 - 2 \sqrt{x - 8}$

$y = - 1 + 2 \sqrt{x - 8}$

Step 2

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

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

Step 3

The standard form is $- 1 - 2 \sqrt{x - 8}, - 1 + 2 \sqrt{x - 8}$ .

$y = - 1 - 2 \sqrt{x - 8}$

$y = - 1 + 2 \sqrt{x - 8}$

Step 4

Write in Standard Form 4x-y^2-2y-33=0

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