Solve the Inequality for x x^2-8x+12<=y

$x^{2} - 8 x + 12 \leq y$

Step 1

Subtract $y$ from both sides of the inequality.

$x^{2} - 8 x + 12 - y \leq 0$

Step 2

Convert the inequality to an equation.

$x^{2} - 8 x + 12 - y = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 1$ , $b = - 8$ , and $c = 12 - y$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot (12 - y))}}{2 \cdot 1}$

Step 5

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot (12 - y)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot (12 - y)}}{2 \cdot 1}$

Apply the distributive property.

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 12 - 4 (- y)}}{2 \cdot 1}$

Multiply $- 4$ by $12$ .

$x = \frac{8 \pm \sqrt{64 - 48 - 4 (- y)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$x = \frac{8 \pm \sqrt{64 - 48 + 4 y}}{2 \cdot 1}$

Subtract $48$ from $64$ .

$x = \frac{8 \pm \sqrt{16 + 4 y}}{2 \cdot 1}$

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

Factor $4$ out of $16$ .

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

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

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

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

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

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

Simplify $\frac{8 \pm 2 \sqrt{4 + y}}{2}$ .

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

Step 6

Consolidate the solutions.

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

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

Solve the Inequality for x x^2-8x+12<=y

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