Solve 6^circ(1+x)^2=72.6

$$6 ^ { \circ } ( 1 + x ) ^ { 2 } = 72.6$$

Solve for x

$$x=\frac{11\sqrt{10}}{10}-1\approx 2.478505426$$

$$x=-\frac{11\sqrt{10}}{10}-1\approx -4.478505426$$

Steps by Finding Square Root

Divide both sides by 6.

$$\frac{6\left(x+1\right)^{2}}{6}=\frac{72.6}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$\left(x+1\right)^{2}=\frac{72.6}{6}$$

Divide 72.6 by 6.

$$\left(x+1\right)^{2}=12.1$$

Take the square root of both sides of the equation.

$$x+1=\frac{11\sqrt{10}}{10} x+1=-\frac{11\sqrt{10}}{10}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{11\sqrt{10}}{10}-1 x+1-1=-\frac{11\sqrt{10}}{10}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{11\sqrt{10}}{10}-1 x=-\frac{11\sqrt{10}}{10}-1$$