Solve 2x^2-3x+c=0

$$2 x ^ { 2 } - 3 x + c = 0$$

Solve for c

$$c=-x\left(2x-3\right)$$

Solution Steps

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

$$-3x+c=-2x^{2}$$

Add 3x to both sides.

$$c=-2x^{2}+3x$$

Solve for x (complex solution)

$$x=\frac{\sqrt{9-8c}+3}{4}$$

$$x=\frac{-\sqrt{9-8c}+3}{4}$$

Solve for x

$$x=\frac{\sqrt{9-8c}+3}{4}$$

$$x=\frac{-\sqrt{9-8c}+3}{4}\text{, }c\leq \frac{9}{8}$$