Solve 1/5(x+3)^2-2=18

$$\frac { 1 } { 5 } ( x + 3 ) ^ { 2 } - 2 = 18$$

Solve for x

$$x=7$$

$$x=-13$$

Steps by Finding Square Root

Add 2 to both sides of the equation.

$$\frac{1}{5}\left(x+3\right)^{2}-2+2=18+2$$

Subtracting 2 from itself leaves 0.

$$\frac{1}{5}\left(x+3\right)^{2}=18+2$$

Add 18 to 2.

$$\frac{1}{5}\left(x+3\right)^{2}=20$$

Multiply both sides by 5.

$$\frac{\frac{1}{5}\left(x+3\right)^{2}}{\frac{1}{5}}=\frac{20}{\frac{1}{5}}$$

Dividing by \frac{1}{5} undoes the multiplication by \frac{1}{5}.

$$\left(x+3\right)^{2}=\frac{20}{\frac{1}{5}}$$

Divide 20 by \frac{1}{5} by multiplying 20 by the reciprocal of \frac{1}{5}.

$$\left(x+3\right)^{2}=100$$

Take the square root of both sides of the equation.

$$x+3=10 x+3=-10$$

Subtract 3 from both sides of the equation.

$$x+3-3=10-3 x+3-3=-10-3$$

Subtracting 3 from itself leaves 0.

$$x=10-3 x=-10-3$$

Subtract 3 from 10.

$$x=7$$

Subtract 3 from -10.

$$x=-13$$

The equation is now solved.

$$x=7 x=-13$$