Solve (n-1)n=500

$$( n - 1 ) n = 500$$

Solve for n

$$n = \frac{\sqrt{2001} + 1}{2} \approx 22.866269246$$

$$n=\frac{1-\sqrt{2001}}{2}\approx -21.866269246$$

Steps Using the Quadratic Formula

Use the distributive property to multiply n-1 by n.

$$n^{2}-n=500$$

Subtract 500 from both sides.

$$n^{2}-n-500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\left(-500\right)}}{2}$$

Multiply -4 times -500.

$$n=\frac{-\left(-1\right)±\sqrt{1+2000}}{2}$$

Add 1 to 2000.

$$n=\frac{-\left(-1\right)±\sqrt{2001}}{2}$$

The opposite of -1 is 1.

$$n=\frac{1±\sqrt{2001}}{2}$$

Now solve the equation n=\frac{1±\sqrt{2001}}{2} when ± is plus. Add 1 to \sqrt{2001}.

$$n=\frac{\sqrt{2001}+1}{2}$$

Now solve the equation n=\frac{1±\sqrt{2001}}{2} when ± is minus. Subtract \sqrt{2001} from 1.

$$n=\frac{1-\sqrt{2001}}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{2001}+1}{2} n=\frac{1-\sqrt{2001}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply n-1 by n.

$$n^{2}-n=500$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-n+\left(-\frac{1}{2}\right)^{2}=500+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-n+\frac{1}{4}=500+\frac{1}{4}$$

Add 500 to \frac{1}{4}.

$$n^{2}-n+\frac{1}{4}=\frac{2001}{4}$$

Factor n^{2}-n+\frac{1}{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}.

$$\left(n-\frac{1}{2}\right)^{2}=\frac{2001}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{2}\right)^{2}}=\sqrt{\frac{2001}{4}}$$

Simplify.

$$n-\frac{1}{2}=\frac{\sqrt{2001}}{2} n-\frac{1}{2}=-\frac{\sqrt{2001}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$n=\frac{\sqrt{2001}+1}{2} n=\frac{1-\sqrt{2001}}{2}$$