Solve 40(1+m)^2=190

$$40 ( 1 + m ) ^ { 2 } = 190$$

Solve for m

$$m=\frac{\sqrt{19}}{2}-1\approx 1.179449472$$

$$m=-\frac{\sqrt{19}}{2}-1\approx -3.179449472$$

Steps by Finding Square Root

Divide both sides by 40.

$$\frac{40\left(m+1\right)^{2}}{40}=\frac{190}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$\left(m+1\right)^{2}=\frac{190}{40}$$

Reduce the fraction \frac{190}{40} to lowest terms by extracting and canceling out 10.

$$\left(m+1\right)^{2}=\frac{19}{4}$$

Take the square root of both sides of the equation.

$$m+1=\frac{\sqrt{19}}{2} m+1=-\frac{\sqrt{19}}{2}$$

Subtract 1 from both sides of the equation.

$$m+1-1=\frac{\sqrt{19}}{2}-1 m+1-1=-\frac{\sqrt{19}}{2}-1$$

Subtracting 1 from itself leaves 0.

$$m=\frac{\sqrt{19}}{2}-1 m=-\frac{\sqrt{19}}{2}-1$$