Solve 11.7500(1+x)^2=10800

$$11.7500 ( 1 + x ) ^ { 2 } = 10800$$

Solve for x

$$x=\frac{120\sqrt{141}}{47}-1\approx 29.317469158$$

$$x=-\frac{120\sqrt{141}}{47}-1\approx -31.317469158$$

Steps by Finding Square Root

Divide both sides of the equation by 11.75, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{11.75\left(x+1\right)^{2}}{11.75}=\frac{10800}{11.75}$$

Dividing by 11.75 undoes the multiplication by 11.75.

$$\left(x+1\right)^{2}=\frac{10800}{11.75}$$

Divide 10800 by 11.75 by multiplying 10800 by the reciprocal of 11.75.

$$\left(x+1\right)^{2}=\frac{43200}{47}$$

Take the square root of both sides of the equation.

$$x+1=\frac{120\sqrt{141}}{47} x+1=-\frac{120\sqrt{141}}{47}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{120\sqrt{141}}{47}-1 x+1-1=-\frac{120\sqrt{141}}{47}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{120\sqrt{141}}{47}-1 x=-\frac{120\sqrt{141}}{47}-1$$