Solve 256left(1+xright)^2=400

$$256 { \left(1+x \right) }^{ 2 } =400$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x = -\frac{9}{4} = -2\frac{1}{4} = -2.25$$

Steps by Finding Square Root

Divide both sides by 256.

$$\frac{256\left(x+1\right)^{2}}{256}=\frac{400}{256}$$

Dividing by 256 undoes the multiplication by 256.

$$\left(x+1\right)^{2}=\frac{400}{256}$$

Reduce the fraction \frac{400}{256} to lowest terms by extracting and canceling out 16.

$$\left(x+1\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$x+1=\frac{5}{4} x+1=-\frac{5}{4}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{5}{4}-1 x+1-1=-\frac{5}{4}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{5}{4}-1 x=-\frac{5}{4}-1$$

Subtract 1 from \frac{5}{4}.

$$x=\frac{1}{4}$$

Subtract 1 from -\frac{5}{4}.

$$x=-\frac{9}{4}$$

The equation is now solved.

$$x=\frac{1}{4} x=-\frac{9}{4}$$