Solve 128(1+x)^2=608

$$128 ( 1 + x ) ^ { 2 } = 608$$

Solve for x

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

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

Steps by Finding Square Root

Divide both sides by 128.

$$\frac{128\left(x+1\right)^{2}}{128}=\frac{608}{128}$$

Dividing by 128 undoes the multiplication by 128.

$$\left(x+1\right)^{2}=\frac{608}{128}$$

Reduce the fraction \frac{608}{128} to lowest terms by extracting and canceling out 32.

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

Take the square root of both sides of the equation.

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

Subtract 1 from both sides of the equation.

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

Subtracting 1 from itself leaves 0.

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