Solve 4/7*(x+4)^2=78

$$\frac { 4 } { 7 } \cdot ( x + 4 ) ^ { 2 } = 78$$

Solve for x

$$x=\frac{\sqrt{546}}{2}-4\approx 7.683321446$$

$$x=-\frac{\sqrt{546}}{2}-4\approx -15.683321446$$

Steps by Finding Square Root

Divide both sides of the equation by \frac{4}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{4}{7}\left(x+4\right)^{2}}{\frac{4}{7}}=\frac{78}{\frac{4}{7}}$$

Dividing by \frac{4}{7} undoes the multiplication by \frac{4}{7}.

$$\left(x+4\right)^{2}=\frac{78}{\frac{4}{7}}$$

Divide 78 by \frac{4}{7} by multiplying 78 by the reciprocal of \frac{4}{7}.

$$\left(x+4\right)^{2}=\frac{273}{2}$$

Take the square root of both sides of the equation.

$$x+4=\frac{\sqrt{546}}{2} x+4=-\frac{\sqrt{546}}{2}$$

Subtract 4 from both sides of the equation.

$$x+4-4=\frac{\sqrt{546}}{2}-4 x+4-4=-\frac{\sqrt{546}}{2}-4$$

Subtracting 4 from itself leaves 0.

$$x=\frac{\sqrt{546}}{2}-4 x=-\frac{\sqrt{546}}{2}-4$$