Solve 3(x+1)^2+4=22

$$3 ( x + 1 ) ^ { 2 } + 4 = 22$$

Solve for x

$$x=\sqrt{6}-1\approx 1.449489743$$

$$x=-\left(\sqrt{6}+1\right)\approx -3.449489743$$

Steps by Finding Square Root

Subtract 4 from both sides of the equation.

$$3\left(x+1\right)^{2}+4-4=22-4$$

Subtracting 4 from itself leaves 0.

$$3\left(x+1\right)^{2}=22-4$$

Subtract 4 from 22.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 18 by 3.

$$\left(x+1\right)^{2}=6$$

Take the square root of both sides of the equation.

$$x+1=\sqrt{6} x+1=-\sqrt{6}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\sqrt{6}-1 x+1-1=-\sqrt{6}-1$$

Subtracting 1 from itself leaves 0.

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$