Solve 45left(1+xright)^2=9

$$45 { \left(1+x \right) }^{ 2 } =9$$

Solve for x

$$x=\frac{\sqrt{5}}{5}-1\approx -0.552786405$$

$$x=-\frac{\sqrt{5}}{5}-1\approx -1.447213595$$

Steps by Finding Square Root

Divide both sides by 45.

$$\frac{45\left(x+1\right)^{2}}{45}=\frac{9}{45}$$

Dividing by 45 undoes the multiplication by 45.

$$\left(x+1\right)^{2}=\frac{9}{45}$$

Reduce the fraction \frac{9}{45} to lowest terms by extracting and canceling out 9.

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

Take the square root of both sides of the equation.

$$x+1=\frac{\sqrt{5}}{5} x+1=-\frac{\sqrt{5}}{5}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{\sqrt{5}}{5}-1 x+1-1=-\frac{\sqrt{5}}{5}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{\sqrt{5}}{5}-1 x=-\frac{\sqrt{5}}{5}-1$$