Solve the Inequality for x y>=2x^2-3x+1

$y \geq 2 x^{2} - 3 x + 1$

Step 1

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

$2 x^{2} - 3 x + 1 \leq y$

Step 2

Subtract $y$ from both sides of the inequality.

$2 x^{2} - 3 x + 1 - y \leq 0$

Step 3

Convert the inequality to an equation.

$2 x^{2} - 3 x + 1 - y = 0$

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

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

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

Step 6

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot (1 - y)}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot (1 - y)}}{2 \cdot 2}$

Apply the distributive property.

$x = \frac{3 \pm \sqrt{9 - 8 \cdot 1 - 8 (- y)}}{2 \cdot 2}$

Multiply $- 8$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 8 - 8 (- y)}}{2 \cdot 2}$

Multiply $- 1$ by $- 8$ .

$x = \frac{3 \pm \sqrt{9 - 8 + 8 y}}{2 \cdot 2}$

Subtract $8$ from $9$ .

$x = \frac{3 \pm \sqrt{1 + 8 y}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm \sqrt{1 + 8 y}}{4}$

Step 7

Consolidate the solutions.

$x = \frac{3 + \sqrt{1 + 8 y}}{4}$

$x = \frac{3 - \sqrt{1 + 8 y}}{4}$