Solve (n-1)^2=2n

$$( n - 1 ) ^ { 2 } = 2 n$$

Solve for n

$$n=\sqrt{3}+2\approx 3.732050808$$

$$n=2-\sqrt{3}\approx 0.267949192$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-1\right)^{2}.

$$n^{2}-2n+1=2n$$

Subtract 2n from both sides.

$$n^{2}-2n+1-2n=0$$

Combine -2n and -2n to get -4n.

$$n^{2}-4n+1=0$$

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

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

Square -4.

$$n=\frac{-\left(-4\right)±\sqrt{16-4}}{2}$$

Add 16 to -4.

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

Take the square root of 12.

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

The opposite of -4 is 4.

$$n=\frac{4±2\sqrt{3}}{2}$$

Now solve the equation n=\frac{4±2\sqrt{3}}{2} when ± is plus. Add 4 to 2\sqrt{3}.

$$n=\frac{2\sqrt{3}+4}{2}$$

Divide 4+2\sqrt{3} by 2.

$$n=\sqrt{3}+2$$

Now solve the equation n=\frac{4±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 4.

$$n=\frac{4-2\sqrt{3}}{2}$$

Divide 4-2\sqrt{3} by 2.

$$n=2-\sqrt{3}$$

The equation is now solved.

$$n=\sqrt{3}+2 n=2-\sqrt{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-1\right)^{2}.

$$n^{2}-2n+1=2n$$

Subtract 2n from both sides.

$$n^{2}-2n+1-2n=0$$

Combine -2n and -2n to get -4n.

$$n^{2}-4n+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$n^{2}-4n=-1$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-4n+\left(-2\right)^{2}=-1+\left(-2\right)^{2}$$

Square -2.

$$n^{2}-4n+4=-1+4$$

Add -1 to 4.

$$n^{2}-4n+4=3$$

Factor n^{2}-4n+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-2\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-2\right)^{2}}=\sqrt{3}$$

Simplify.

$$n-2=\sqrt{3} n-2=-\sqrt{3}$$

Add 2 to both sides of the equation.

$$n=\sqrt{3}+2 n=2-\sqrt{3}$$