Solve Using the Quadratic Formula x^2-y=-4x-3

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

Move all the expressions to the left side of the equation.

Add $4 x$ to both sides of the equation.

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

Add $3$ to both sides of the equation.

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

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 $c = - y + 3$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

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

Multiply $- 4$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $- 4$ .

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

Multiply $- 4$ by $3$ .

$x = \frac{- 4 \pm \sqrt{16 + 4 y - 12}}{2 \cdot 1}$

Subtract $12$ from $16$ .

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

Factor $4$ out of $4 y + 4$ .

Factor $4$ out of $4 y$ .

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

Factor $4$ out of $4$ .

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

Factor $4$ out of $4 (y) + 4 (1)$ .

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

Rewrite $4 (y + 1)$ as $2^{2} (y + 1^{2})$ .

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

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

Rewrite $1$ as $1^{2}$ .

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

Pull terms out from under the radical.

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

One to any power is one.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 4 \pm 2 \sqrt{y + 1}}{2}$ .

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