Solve (n-3)^2+n^2=(n+6)^2

$$( n - 3 ) ^ { 2 } + n ^ { 2 } = ( n + 6 ) ^ { 2 }$$

Solve for n

$$n=6\sqrt{3}+9\approx 19.392304845$$

$$n=9-6\sqrt{3}\approx -1.392304845$$

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}.

$$n^{2}-6n+9+n^{2}=\left(n+6\right)^{2}$$

Combine n^{2} and n^{2} to get 2n^{2}.

$$2n^{2}-6n+9=\left(n+6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n+6\right)^{2}.

$$2n^{2}-6n+9=n^{2}+12n+36$$

Subtract n^{2} from both sides.

$$2n^{2}-6n+9-n^{2}=12n+36$$

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

$$n^{2}-6n+9=12n+36$$

Subtract 12n from both sides.

$$n^{2}-6n+9-12n=36$$

Combine -6n and -12n to get -18n.

$$n^{2}-18n+9=36$$

Subtract 36 from both sides.

$$n^{2}-18n+9-36=0$$

Subtract 36 from 9 to get -27.

$$n^{2}-18n-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-27\right)}}{2}$$

Square -18.

$$n=\frac{-\left(-18\right)±\sqrt{324-4\left(-27\right)}}{2}$$

Multiply -4 times -27.

$$n=\frac{-\left(-18\right)±\sqrt{324+108}}{2}$$

Add 324 to 108.

$$n=\frac{-\left(-18\right)±\sqrt{432}}{2}$$

Take the square root of 432.

$$n=\frac{-\left(-18\right)±12\sqrt{3}}{2}$$

The opposite of -18 is 18.

$$n=\frac{18±12\sqrt{3}}{2}$$

Now solve the equation n=\frac{18±12\sqrt{3}}{2} when ± is plus. Add 18 to 12\sqrt{3}.

$$n=\frac{12\sqrt{3}+18}{2}$$

Divide 18+12\sqrt{3} by 2.

$$n=6\sqrt{3}+9$$

Now solve the equation n=\frac{18±12\sqrt{3}}{2} when ± is minus. Subtract 12\sqrt{3} from 18.

$$n=\frac{18-12\sqrt{3}}{2}$$

Divide 18-12\sqrt{3} by 2.

$$n=9-6\sqrt{3}$$

The equation is now solved.

$$n=6\sqrt{3}+9 n=9-6\sqrt{3}$$

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}.

$$n^{2}-6n+9+n^{2}=\left(n+6\right)^{2}$$

Combine n^{2} and n^{2} to get 2n^{2}.

$$2n^{2}-6n+9=\left(n+6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n+6\right)^{2}.

$$2n^{2}-6n+9=n^{2}+12n+36$$

Subtract n^{2} from both sides.

$$2n^{2}-6n+9-n^{2}=12n+36$$

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

$$n^{2}-6n+9=12n+36$$

Subtract 12n from both sides.

$$n^{2}-6n+9-12n=36$$

Combine -6n and -12n to get -18n.

$$n^{2}-18n+9=36$$

Subtract 9 from both sides.

$$n^{2}-18n=36-9$$

Subtract 9 from 36 to get 27.

$$n^{2}-18n=27$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-18n+\left(-9\right)^{2}=27+\left(-9\right)^{2}$$

Square -9.

$$n^{2}-18n+81=27+81$$

Add 27 to 81.

$$n^{2}-18n+81=108$$

Factor n^{2}-18n+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-9\right)^{2}=108$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-9\right)^{2}}=\sqrt{108}$$

Simplify.

$$n-9=6\sqrt{3} n-9=-6\sqrt{3}$$

Add 9 to both sides of the equation.

$$n=6\sqrt{3}+9 n=9-6\sqrt{3}$$