Solve 2d^2+4c-5=0

$$2 d ^ { 2 } + 4 c - 5 = 0$$

Solve for c

$$c=-\frac{d^{2}}{2}+\frac{5}{4}$$

Steps for Solving Linear Equation

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

$$4c-5=-2d^{2}$$

Add 5 to both sides.

$$4c=-2d^{2}+5$$

The equation is in standard form.

$$4c=5-2d^{2}$$

Divide both sides by 4.

$$\frac{4c}{4}=\frac{5-2d^{2}}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$c=\frac{5-2d^{2}}{4}$$

Divide -2d^{2}+5 by 4.

$$c=-\frac{d^{2}}{2}+\frac{5}{4}$$

Solve for d (complex solution)

$$d=-\frac{\sqrt{10-8c}}{2}$$

$$d=\frac{\sqrt{10-8c}}{2}$$

Solve for d

$$d=\frac{\sqrt{10-8c}}{2}$$

$$d=-\frac{\sqrt{10-8c}}{2}\text{, }c\leq \frac{5}{4}$$