Solve -12sqrt{-c-2}+16=4

$$- 12 \sqrt { - c - 2 } + 16 = 4$$

Solve for c

$$c=-3$$

Steps for Solving Radical Equations

Subtract 16 from both sides of the equation.

$$-12\sqrt{-c-2}+16-16=4-16$$

Subtracting 16 from itself leaves 0.

$$-12\sqrt{-c-2}=4-16$$

Subtract 16 from 4.

$$-12\sqrt{-c-2}=-12$$

Divide both sides by -12.

$$\frac{-12\sqrt{-c-2}}{-12}=-\frac{12}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$\sqrt{-c-2}=-\frac{12}{-12}$$

Divide -12 by -12.

$$\sqrt{-c-2}=1$$

Square both sides of the equation.

$$-c-2=1$$

Add 2 to both sides of the equation.

$$-c-2-\left(-2\right)=1-\left(-2\right)$$

Subtracting -2 from itself leaves 0.

$$-c=1-\left(-2\right)$$

Subtract -2 from 1.

$$-c=3$$

Divide both sides by -1.

$$\frac{-c}{-1}=\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$c=\frac{3}{-1}$$

Divide 3 by -1.

$$c=-3$$