Solve 0=n/2left(36+(n-1)-2right)

$$0 = \frac{ n }{ 2 } \left( 36+(n-1)-2 \right)$$

Solve for n

$$n=-33$$

$$n=0$$

Steps Using Factoring

Multiply both sides of the equation by 2.

$$0=n\left(36+n-1-2\right)$$

Subtract 1 from 36 to get 35.

$$0=n\left(35+n-2\right)$$

Subtract 2 from 35 to get 33.

$$0=n\left(33+n\right)$$

Use the distributive property to multiply n by 33+n.

$$0=33n+n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$33n+n^{2}=0$$

Factor out n.

$$n\left(33+n\right)=0$$

To find equation solutions, solve n=0 and 33+n=0.

$$n=0 n=-33$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$0=n\left(36+n-1-2\right)$$

Subtract 1 from 36 to get 35.

$$0=n\left(35+n-2\right)$$

Subtract 2 from 35 to get 33.

$$0=n\left(33+n\right)$$

Use the distributive property to multiply n by 33+n.

$$0=33n+n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$33n+n^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$n^{2}+33n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 33 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-33±\sqrt{33^{2}}}{2}$$

Take the square root of 33^{2}.

$$n=\frac{-33±33}{2}$$

Now solve the equation n=\frac{-33±33}{2} when ± is plus. Add -33 to 33.

$$n=\frac{0}{2}$$

Divide 0 by 2.

$$n=0$$

Now solve the equation n=\frac{-33±33}{2} when ± is minus. Subtract 33 from -33.

$$n=-\frac{66}{2}$$

Divide -66 by 2.

$$n=-33$$

The equation is now solved.

$$n=0 n=-33$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$0=n\left(36+n-1-2\right)$$

Subtract 1 from 36 to get 35.

$$0=n\left(35+n-2\right)$$

Subtract 2 from 35 to get 33.

$$0=n\left(33+n\right)$$

Use the distributive property to multiply n by 33+n.

$$0=33n+n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$33n+n^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$n^{2}+33n=0$$

Divide 33, the coefficient of the x term, by 2 to get \frac{33}{2}. Then add the square of \frac{33}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+33n+\left(\frac{33}{2}\right)^{2}=\left(\frac{33}{2}\right)^{2}$$

Square \frac{33}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+33n+\frac{1089}{4}=\frac{1089}{4}$$

Factor n^{2}+33n+\frac{1089}{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}.

$$\left(n+\frac{33}{2}\right)^{2}=\frac{1089}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{33}{2}\right)^{2}}=\sqrt{\frac{1089}{4}}$$

Simplify.

$$n+\frac{33}{2}=\frac{33}{2} n+\frac{33}{2}=-\frac{33}{2}$$

Subtract \frac{33}{2} from both sides of the equation.

$$n=0 n=-33$$