Solve n/2[n+1]=100

$$\frac { n } { 2 } [ n + 1 ] = 100$$

Solve for n

$$n = \frac{3 \sqrt{89} - 1}{2} \approx 13.650971698$$

$$n=\frac{-3\sqrt{89}-1}{2}\approx -14.650971698$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$n\left(n+1\right)=200$$

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

$$n^{2}+n=200$$

Subtract 200 from both sides.

$$n^{2}+n-200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-1±\sqrt{1^{2}-4\left(-200\right)}}{2}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\left(-200\right)}}{2}$$

Multiply -4 times -200.

$$n=\frac{-1±\sqrt{1+800}}{2}$$

Add 1 to 800.

$$n=\frac{-1±\sqrt{801}}{2}$$

Take the square root of 801.

$$n=\frac{-1±3\sqrt{89}}{2}$$

Now solve the equation n=\frac{-1±3\sqrt{89}}{2} when ± is plus. Add -1 to 3\sqrt{89}.

$$n=\frac{3\sqrt{89}-1}{2}$$

Now solve the equation n=\frac{-1±3\sqrt{89}}{2} when ± is minus. Subtract 3\sqrt{89} from -1.

$$n=\frac{-3\sqrt{89}-1}{2}$$

The equation is now solved.

$$n=\frac{3\sqrt{89}-1}{2} n=\frac{-3\sqrt{89}-1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$n\left(n+1\right)=200$$

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

$$n^{2}+n=200$$

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.

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=200+\left(\frac{1}{2}\right)^{2}$$

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

$$n^{2}+n+\frac{1}{4}=200+\frac{1}{4}$$

Add 200 to \frac{1}{4}.

$$n^{2}+n+\frac{1}{4}=\frac{801}{4}$$

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}.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{801}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{801}{4}}$$

Simplify.

$$n+\frac{1}{2}=\frac{3\sqrt{89}}{2} n+\frac{1}{2}=-\frac{3\sqrt{89}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$n=\frac{3\sqrt{89}-1}{2} n=\frac{-3\sqrt{89}-1}{2}$$