Solve y=-sqrt{-3x-4}

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

Solve for x

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

$$y\leq 0$$

Steps for Solving Radical Equations

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

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

Divide both sides by -1.

$$\frac{-\sqrt{-3x-4}}{-1}=\frac{y}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$\sqrt{-3x-4}=\frac{y}{-1}$$

Divide y by -1.

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

Square both sides of the equation.

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

Add 4 to both sides of the equation.

$$-3x-4-\left(-4\right)=y^{2}-\left(-4\right)$$

Subtracting -4 from itself leaves 0.

$$-3x=y^{2}-\left(-4\right)$$

Subtract -4 from y^{2}.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

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

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

Solve for x (complex solution)

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

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

Solve for y

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

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