Solve for x y>=x^2-2x-3

$y \geq x^{2} - 2 x - 3$

Rewrite so $x$ is on the left side of the inequality.

$x^{2} - 2 x - 3 \leq y$

Subtract $y$ from both sides of the inequality.

$x^{2} - 2 x - 3 - y \leq 0$

Convert the inequality to an equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 3 - y$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 4$ by $- 3$ .

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

Multiply $- 1$ by $- 4$ .

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

Add $4$ and $12$ .

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

Factor $4$ out of $16 + 4 y$ .

Factor $4$ out of $16$ .

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

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

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

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

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

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

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

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

Consolidate the solutions.

$x = 1 + \sqrt{4 + y}$

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

Solve for x y>=x^2-2x-3

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