Solve (r+5)^2=2r^2

$$( r + 5 ) ^ { 2 } = 2 r ^ { 2 }$$

Solve for r

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

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

Steps Using the Quadratic Formula

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

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

Subtract 2r^{2} from both sides.

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

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

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

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}.

$$r=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\times 25}}{2\left(-1\right)}$$

Square 10.

$$r=\frac{-10±\sqrt{100-4\left(-1\right)\times 25}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$r=\frac{-10±\sqrt{100+4\times 25}}{2\left(-1\right)}$$

Multiply 4 times 25.

$$r=\frac{-10±\sqrt{100+100}}{2\left(-1\right)}$$

Add 100 to 100.

$$r=\frac{-10±\sqrt{200}}{2\left(-1\right)}$$

Take the square root of 200.

$$r=\frac{-10±10\sqrt{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$r=\frac{-10±10\sqrt{2}}{-2}$$

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.

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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

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

Subtract 2r^{2} from both sides.

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

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

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

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

$$\frac{-r^{2}+10r}{-1}=-\frac{25}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$r^{2}+\frac{10}{-1}r=-\frac{25}{-1}$$

Divide 10 by -1.

$$r^{2}-10r=-\frac{25}{-1}$$

Divide -25 by -1.

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

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.

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