Solve 200=n(3n+1)/2

$$200 = \frac { n ( 3 n + 1 ) } { 2 }$$

Solve for n

$$n = \frac{\sqrt{4801} - 1}{6} \approx 11.381541468$$

$$n=\frac{-\sqrt{4801}-1}{6}\approx -11.714874801$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$200\times 2=n\left(3n+1\right)$$

Multiply 200 and 2 to get 400.

$$400=n\left(3n+1\right)$$

Use the distributive property to multiply n by 3n+1.

$$400=3n^{2}+n$$

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

$$3n^{2}+n=400$$

Subtract 400 from both sides.

$$3n^{2}+n-400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and -400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-1±\sqrt{1^{2}-4\times 3\left(-400\right)}}{2\times 3}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\times 3\left(-400\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-1±\sqrt{1-12\left(-400\right)}}{2\times 3}$$

Multiply -12 times -400.

$$n=\frac{-1±\sqrt{1+4800}}{2\times 3}$$

Add 1 to 4800.

$$n=\frac{-1±\sqrt{4801}}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-1±\sqrt{4801}}{6}$$

Now solve the equation n=\frac{-1±\sqrt{4801}}{6} when ± is plus. Add -1 to \sqrt{4801}.

$$n=\frac{\sqrt{4801}-1}{6}$$

Now solve the equation n=\frac{-1±\sqrt{4801}}{6} when ± is minus. Subtract \sqrt{4801} from -1.

$$n=\frac{-\sqrt{4801}-1}{6}$$

The equation is now solved.

$$n=\frac{\sqrt{4801}-1}{6} n=\frac{-\sqrt{4801}-1}{6}$$

Steps for Completing the Square

Multiply both sides by 2.

$$200\times 2=n\left(3n+1\right)$$

Multiply 200 and 2 to get 400.

$$400=n\left(3n+1\right)$$

Use the distributive property to multiply n by 3n+1.

$$400=3n^{2}+n$$

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

$$3n^{2}+n=400$$

Divide both sides by 3.

$$\frac{3n^{2}+n}{3}=\frac{400}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$n^{2}+\frac{1}{3}n=\frac{400}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{1}{3}n+\left(\frac{1}{6}\right)^{2}=\frac{400}{3}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{1}{3}n+\frac{1}{36}=\frac{400}{3}+\frac{1}{36}$$

Add \frac{400}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{1}{3}n+\frac{1}{36}=\frac{4801}{36}$$

Factor n^{2}+\frac{1}{3}n+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{1}{6}\right)^{2}=\frac{4801}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{6}\right)^{2}}=\sqrt{\frac{4801}{36}}$$

Simplify.

$$n+\frac{1}{6}=\frac{\sqrt{4801}}{6} n+\frac{1}{6}=-\frac{\sqrt{4801}}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$n=\frac{\sqrt{4801}-1}{6} n=\frac{-\sqrt{4801}-1}{6}$$