Solve 1280(1+x)^2=2880

$$1280 ( 1 + x ) ^ { 2 } = 2880$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps by Finding Square Root

Divide both sides by 1280.

$$\frac{1280\left(x+1\right)^{2}}{1280}=\frac{2880}{1280}$$

Dividing by 1280 undoes the multiplication by 1280.

$$\left(x+1\right)^{2}=\frac{2880}{1280}$$

Reduce the fraction \frac{2880}{1280} to lowest terms by extracting and canceling out 320.

$$\left(x+1\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$x+1=\frac{3}{2} x+1=-\frac{3}{2}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{3}{2}-1 x+1-1=-\frac{3}{2}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{3}{2}-1 x=-\frac{3}{2}-1$$

Subtract 1 from \frac{3}{2}.

$$x=\frac{1}{2}$$

Subtract 1 from -\frac{3}{2}.

$$x=-\frac{5}{2}$$

The equation is now solved.

$$x=\frac{1}{2} x=-\frac{5}{2}$$