Solve 4(x+1)^2=8

$$4 ( x + 1 ) ^ { 2 } = 8$$

Solve for x

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\left(\sqrt{2}+1\right)\approx -2.414213562$$

Steps by Finding Square Root

Divide both sides by 4.

$$\frac{4\left(x+1\right)^{2}}{4}=\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$\left(x+1\right)^{2}=\frac{8}{4}$$

Divide 8 by 4.

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

Take the square root of both sides of the equation.

$$x+1=\sqrt{2} x+1=-\sqrt{2}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\sqrt{2}-1 x+1-1=-\sqrt{2}-1$$

Subtracting 1 from itself leaves 0.

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