Solve 2y^2-4x+1=0

$$2 y ^ { 2 } - 4 x + 1 = 0$$

Solve for x

$$x=\frac{y^{2}}{2}+\frac{1}{4}$$

Steps for Solving Linear Equation

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

$$-4x+1=-2y^{2}$$

Subtract 1 from both sides.

$$-4x=-2y^{2}-1$$

Divide both sides by -4.

$$\frac{-4x}{-4}=\frac{-2y^{2}-1}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x=\frac{-2y^{2}-1}{-4}$$

Divide -2y^{2}-1 by -4.

$$x=\frac{y^{2}}{2}+\frac{1}{4}$$

Solve for y (complex solution)

$$y=-\frac{\sqrt{8x-2}}{2}$$

$$y=\frac{\sqrt{8x-2}}{2}$$

Solve for y

$$y=\frac{\sqrt{8x-2}}{2}$$

$$y=-\frac{\sqrt{8x-2}}{2}\text{, }x\geq \frac{1}{4}$$