Solve left(1+xright)^2*40=135

$${ \left(1+x \right) }^{ 2 } \times 40=135$$

Solve for x

$$x=\frac{3\sqrt{6}}{4}-1\approx 0.837117307$$

$$x=-\frac{3\sqrt{6}}{4}-1\approx -2.837117307$$

Steps by Finding Square Root

Divide both sides by 40.

$$\frac{40\left(x+1\right)^{2}}{40}=\frac{135}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$\left(x+1\right)^{2}=\frac{135}{40}$$

Reduce the fraction \frac{135}{40} to lowest terms by extracting and canceling out 5.

$$\left(x+1\right)^{2}=\frac{27}{8}$$

Take the square root of both sides of the equation.

$$x+1=\frac{3\sqrt{6}}{4} x+1=-\frac{3\sqrt{6}}{4}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{3\sqrt{6}}{4}-1 x+1-1=-\frac{3\sqrt{6}}{4}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{3\sqrt{6}}{4}-1 x=-\frac{3\sqrt{6}}{4}-1$$