Solve 3(x+2)^2+6=21

$$3 ( x + 2 ) ^ { 2 } + 6 = 21$$

Solve for x

$$x=\sqrt{5}-2\approx 0.236067977$$

$$x=-\left(\sqrt{5}+2\right)\approx -4.236067977$$

Steps by Finding Square Root

Subtract 6 from both sides of the equation.

$$3\left(x+2\right)^{2}+6-6=21-6$$

Subtracting 6 from itself leaves 0.

$$3\left(x+2\right)^{2}=21-6$$

Subtract 6 from 21.

$$3\left(x+2\right)^{2}=15$$

Divide both sides by 3.

$$\frac{3\left(x+2\right)^{2}}{3}=\frac{15}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$\left(x+2\right)^{2}=\frac{15}{3}$$

Divide 15 by 3.

$$\left(x+2\right)^{2}=5$$

Take the square root of both sides of the equation.

$$x+2=\sqrt{5} x+2=-\sqrt{5}$$

Subtract 2 from both sides of the equation.

$$x+2-2=\sqrt{5}-2 x+2-2=-\sqrt{5}-2$$

Subtracting 2 from itself leaves 0.

$$x=\sqrt{5}-2 x=-\sqrt{5}-2$$