Solve 2x^2-5x+h=0

$$2 x ^ { 2 } - 5 x + h = 0$$

Solve for h

$$h=-x\left(2x-5\right)$$

Solution Steps

Subtract 2x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-5x+h=-2x^{2}$$

Add 5x to both sides.

$$h=-2x^{2}+5x$$

Solve for x (complex solution)

$$x=\frac{\sqrt{25-8h}+5}{4}$$

$$x=\frac{-\sqrt{25-8h}+5}{4}$$

Solve for x

$$x=\frac{\sqrt{25-8h}+5}{4}$$

$$x=\frac{-\sqrt{25-8h}+5}{4}\text{, }h\leq \frac{25}{8}$$