Solve pir^2=95

$$\pi r ^ { 2 } = 95$$

Solve for r

$$r=\sqrt{\frac{95}{\pi }}\approx 5.499039842$$

$$r=-\sqrt{\frac{95}{\pi }}\approx -5.499039842$$

Steps by Finding Square Root

Divide both sides by \pi .

$$\frac{\pi r^{2}}{\pi }=\frac{95}{\pi }$$

Dividing by \pi undoes the multiplication by \pi .

$$r^{2}=\frac{95}{\pi }$$

Take the square root of both sides of the equation.

$$r=\frac{95}{\sqrt{95\pi }} r=-\frac{95}{\sqrt{95\pi }}$$

Steps Using the Quadratic Formula

Subtract 95 from both sides.

$$\pi r^{2}-95=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \pi for a, 0 for b, and -95 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{0±\sqrt{0^{2}-4\pi \left(-95\right)}}{2\pi }$$

Square 0.

$$r=\frac{0±\sqrt{-4\pi \left(-95\right)}}{2\pi }$$

Multiply -4 times \pi .

$$r=\frac{0±\sqrt{\left(-4\pi \right)\left(-95\right)}}{2\pi }$$

Multiply -4\pi times -95.

$$r=\frac{0±\sqrt{380\pi }}{2\pi }$$

Take the square root of 380\pi .

$$r=\frac{0±2\sqrt{95\pi }}{2\pi }$$

Now solve the equation r=\frac{0±2\sqrt{95\pi }}{2\pi } when ± is plus.

$$r=\frac{95}{\sqrt{95\pi }}$$

Now solve the equation r=\frac{0±2\sqrt{95\pi }}{2\pi } when ± is minus.

$$r=-\frac{95}{\sqrt{95\pi }}$$

The equation is now solved.

$$r=\frac{95}{\sqrt{95\pi }} r=-\frac{95}{\sqrt{95\pi }}$$