Solve 2y^2+n=0

$$2 y ^ { 2 } + n = 0$$

Solve for n

$$n=-2y^{2}$$

Solution Steps

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

$$n=-2y^{2}$$

Solve for y (complex solution)

$$y=-\frac{i\sqrt{2n}}{2}$$

$$y=\frac{i\sqrt{2n}}{2}$$

Solve for y

$$y=\frac{\sqrt{-2n}}{2}$$

$$y=-\frac{\sqrt{-2n}}{2}\text{, }n\leq 0$$