Solve y=sqrt{3-4x}

$$y = \sqrt { 3 - 4 x }$$

Solve for x

$$x=\frac{3-y^{2}}{4}$$

$$y\geq 0$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

$$\sqrt{3-4x}=y$$

Square both sides of the equation.

$$-4x+3=y^{2}$$

Subtract 3 from both sides of the equation.

$$-4x+3-3=y^{2}-3$$

Subtracting 3 from itself leaves 0.

$$-4x=y^{2}-3$$

Divide both sides by -4.

$$\frac{-4x}{-4}=\frac{y^{2}-3}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x=\frac{y^{2}-3}{-4}$$

Divide y^{2}-3 by -4.

$$x=\frac{3-y^{2}}{4}$$

Solve for x (complex solution)

$$x=\frac{3-y^{2}}{4}$$

$$arg(y)<\pi \text{ or }y=0$$

Solve for y

$$y=\sqrt{3-4x}$$

$$x\leq \frac{3}{4}$$