Solve 40+40(1+x)^2=135

$$40 + 40 ( 1 + x ) ^ { 2 } = 135$$

Solve for x

$$x=\frac{\sqrt{38}}{4}-1\approx 0.541103501$$

$$x=-\frac{\sqrt{38}}{4}-1\approx -2.541103501$$

Steps by Finding Square Root

Subtract 40 from both sides of the equation.

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

Subtracting 40 from itself leaves 0.

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

Subtract 40 from 135.

$$40\left(x+1\right)^{2}=95$$

Divide both sides by 40.

$$\frac{40\left(x+1\right)^{2}}{40}=\frac{95}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$\left(x+1\right)^{2}=\frac{95}{40}$$

Reduce the fraction \frac{95}{40} to lowest terms by extracting and canceling out 5.

$$\left(x+1\right)^{2}=\frac{19}{8}$$

Take the square root of both sides of the equation.

$$x+1=\frac{\sqrt{38}}{4} x+1=-\frac{\sqrt{38}}{4}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{\sqrt{38}}{4}-1 x+1-1=-\frac{\sqrt{38}}{4}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{\sqrt{38}}{4}-1 x=-\frac{\sqrt{38}}{4}-1$$