Solve (n-5)^2=2n

$$( n - 5 ) ^ { 2 } = 2 n$$

Solve for n

$$n=\sqrt{11}+6\approx 9.31662479$$

$$n=6-\sqrt{11}\approx 2.68337521$$

Steps Using the Quadratic Formula

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

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

Subtract 2n from both sides.

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

Combine -10n and -2n to get -12n.

$$n^{2}-12n+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 25}}{2}$$

Square -12.

$$n=\frac{-\left(-12\right)±\sqrt{144-4\times 25}}{2}$$

Multiply -4 times 25.

$$n=\frac{-\left(-12\right)±\sqrt{144-100}}{2}$$

Add 144 to -100.

$$n=\frac{-\left(-12\right)±\sqrt{44}}{2}$$

Take the square root of 44.

$$n=\frac{-\left(-12\right)±2\sqrt{11}}{2}$$

The opposite of -12 is 12.

$$n=\frac{12±2\sqrt{11}}{2}$$

Now solve the equation n=\frac{12±2\sqrt{11}}{2} when ± is plus. Add 12 to 2\sqrt{11}.

$$n=\frac{2\sqrt{11}+12}{2}$$

Divide 12+2\sqrt{11} by 2.

$$n=\sqrt{11}+6$$

Now solve the equation n=\frac{12±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from 12.

$$n=\frac{12-2\sqrt{11}}{2}$$

Divide 12-2\sqrt{11} by 2.

$$n=6-\sqrt{11}$$

The equation is now solved.

$$n=\sqrt{11}+6 n=6-\sqrt{11}$$

Steps for Completing the Square

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

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

Subtract 2n from both sides.

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

Combine -10n and -2n to get -12n.

$$n^{2}-12n+25=0$$

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

$$n^{2}-12n=-25$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-12n+\left(-6\right)^{2}=-25+\left(-6\right)^{2}$$

Square -6.

$$n^{2}-12n+36=-25+36$$

Add -25 to 36.

$$n^{2}-12n+36=11$$

Factor n^{2}-12n+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-6\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-6\right)^{2}}=\sqrt{11}$$

Simplify.

$$n-6=\sqrt{11} n-6=-\sqrt{11}$$

Add 6 to both sides of the equation.

$$n=\sqrt{11}+6 n=6-\sqrt{11}$$