Solve {(pir^2)}*2=pi(r+5)^2

$${( \pi r ^ { 2 } )} \times 2 = \pi {( r + 5 )} ^ { 2 }$$

Solve for r

$$r=5\sqrt{2}+5\approx 12.071067812$$

$$r=5-5\sqrt{2}\approx -2.071067812$$

Steps Using the Quadratic Formula

Cancel out \pi on both sides.

$$r^{2}\times 2=\left(r+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(r+5\right)^{2}.

$$r^{2}\times 2=r^{2}+10r+25$$

Subtract r^{2} from both sides.

$$r^{2}\times 2-r^{2}=10r+25$$

Combine r^{2}\times 2 and -r^{2} to get r^{2}.

$$r^{2}=10r+25$$

Subtract 10r from both sides.

$$r^{2}-10r=25$$

Subtract 25 from both sides.

$$r^{2}-10r-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-25\right)}}{2}$$

Square -10.

$$r=\frac{-\left(-10\right)±\sqrt{100-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

$$r=\frac{-\left(-10\right)±\sqrt{100+100}}{2}$$

Add 100 to 100.

$$r=\frac{-\left(-10\right)±\sqrt{200}}{2}$$

Take the square root of 200.

$$r=\frac{-\left(-10\right)±10\sqrt{2}}{2}$$

The opposite of -10 is 10.

$$r=\frac{10±10\sqrt{2}}{2}$$

Now solve the equation r=\frac{10±10\sqrt{2}}{2} when ± is plus. Add 10 to 10\sqrt{2}.

$$r=\frac{10\sqrt{2}+10}{2}$$

Divide 10+10\sqrt{2} by 2.

$$r=5\sqrt{2}+5$$

Now solve the equation r=\frac{10±10\sqrt{2}}{2} when ± is minus. Subtract 10\sqrt{2} from 10.

$$r=\frac{10-10\sqrt{2}}{2}$$

Divide 10-10\sqrt{2} by 2.

$$r=5-5\sqrt{2}$$

The equation is now solved.

$$r=5\sqrt{2}+5 r=5-5\sqrt{2}$$

Steps for Completing the Square

Cancel out \pi on both sides.

$$r^{2}\times 2=\left(r+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(r+5\right)^{2}.

$$r^{2}\times 2=r^{2}+10r+25$$

Subtract r^{2} from both sides.

$$r^{2}\times 2-r^{2}=10r+25$$

Combine r^{2}\times 2 and -r^{2} to get r^{2}.

$$r^{2}=10r+25$$

Subtract 10r from both sides.

$$r^{2}-10r=25$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-10r+\left(-5\right)^{2}=25+\left(-5\right)^{2}$$

Square -5.

$$r^{2}-10r+25=25+25$$

Add 25 to 25.

$$r^{2}-10r+25=50$$

Factor r^{2}-10r+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-5\right)^{2}=50$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-5\right)^{2}}=\sqrt{50}$$

Simplify.

$$r-5=5\sqrt{2} r-5=-5\sqrt{2}$$

Add 5 to both sides of the equation.

$$r=5\sqrt{2}+5 r=5-5\sqrt{2}$$