Solve 4pir^2=12pi

$$4 \pi r ^ { 2 } = 12 \pi$$

Solve for r

$$r=\sqrt{3}\approx 1.732050808$$

$$r=-\sqrt{3}\approx -1.732050808$$

Steps by Finding Square Root

Cancel out \pi on both sides.

$$4r^{2}=12$$

Divide both sides by 4.

$$r^{2}=\frac{12}{4}$$

Divide 12 by 4 to get 3.

$$r^{2}=3$$

Take the square root of both sides of the equation.

$$r=\sqrt{3} r=-\sqrt{3}$$

Steps Using the Quadratic Formula

Cancel out \pi on both sides.

$$4r^{2}=12$$

Divide both sides by 4.

$$r^{2}=\frac{12}{4}$$

Divide 12 by 4 to get 3.

$$r^{2}=3$$

Subtract 3 from both sides.

$$r^{2}-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{0±\sqrt{0^{2}-4\left(-3\right)}}{2}$$

Square 0.

$$r=\frac{0±\sqrt{-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$r=\frac{0±\sqrt{12}}{2}$$

Take the square root of 12.

$$r=\frac{0±2\sqrt{3}}{2}$$

Now solve the equation r=\frac{0±2\sqrt{3}}{2} when ± is plus.

$$r=\sqrt{3}$$

Now solve the equation r=\frac{0±2\sqrt{3}}{2} when ± is minus.

$$r=-\sqrt{3}$$

The equation is now solved.

$$r=\sqrt{3} r=-\sqrt{3}$$