Solve (n-3)^2=2n

$$( n - 3 ) ^ { 2 } = 2 n$$

Solve for n

$$n=\sqrt{7}+4\approx 6.645751311$$

$$n=4-\sqrt{7}\approx 1.354248689$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-3\right)^{2}.

$$n^{2}-6n+9=2n$$

Subtract 2n from both sides.

$$n^{2}-6n+9-2n=0$$

Combine -6n and -2n to get -8n.

$$n^{2}-8n+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 9}}{2}$$

Square -8.

$$n=\frac{-\left(-8\right)±\sqrt{64-4\times 9}}{2}$$

Multiply -4 times 9.

$$n=\frac{-\left(-8\right)±\sqrt{64-36}}{2}$$

Add 64 to -36.

$$n=\frac{-\left(-8\right)±\sqrt{28}}{2}$$

Take the square root of 28.

$$n=\frac{-\left(-8\right)±2\sqrt{7}}{2}$$

The opposite of -8 is 8.

$$n=\frac{8±2\sqrt{7}}{2}$$

Now solve the equation n=\frac{8±2\sqrt{7}}{2} when ± is plus. Add 8 to 2\sqrt{7}.

$$n=\frac{2\sqrt{7}+8}{2}$$

Divide 8+2\sqrt{7} by 2.

$$n=\sqrt{7}+4$$

Now solve the equation n=\frac{8±2\sqrt{7}}{2} when ± is minus. Subtract 2\sqrt{7} from 8.

$$n=\frac{8-2\sqrt{7}}{2}$$

Divide 8-2\sqrt{7} by 2.

$$n=4-\sqrt{7}$$

The equation is now solved.

$$n=\sqrt{7}+4 n=4-\sqrt{7}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-3\right)^{2}.

$$n^{2}-6n+9=2n$$

Subtract 2n from both sides.

$$n^{2}-6n+9-2n=0$$

Combine -6n and -2n to get -8n.

$$n^{2}-8n+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$n^{2}-8n=-9$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-8n+\left(-4\right)^{2}=-9+\left(-4\right)^{2}$$

Square -4.

$$n^{2}-8n+16=-9+16$$

Add -9 to 16.

$$n^{2}-8n+16=7$$

Factor n^{2}-8n+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-4\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-4\right)^{2}}=\sqrt{7}$$

Simplify.

$$n-4=\sqrt{7} n-4=-\sqrt{7}$$

Add 4 to both sides of the equation.

$$n=\sqrt{7}+4 n=4-\sqrt{7}$$