Solve 13320=25n^2+25n

$$13320 = 25 n ^ { 2 } + 25 n$$

Solve for n

$$n=\frac{\sqrt{53305}}{10}-\frac{1}{2}\approx 22.587875606$$

$$n=-\frac{\sqrt{53305}}{10}-\frac{1}{2}\approx -23.587875606$$

Steps Using the Quadratic Formula

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

$$25n^{2}+25n=13320$$

Subtract 13320 from both sides.

$$25n^{2}+25n-13320=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 25 for b, and -13320 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-25±\sqrt{25^{2}-4\times 25\left(-13320\right)}}{2\times 25}$$

Square 25.

$$n=\frac{-25±\sqrt{625-4\times 25\left(-13320\right)}}{2\times 25}$$

Multiply -4 times 25.

$$n=\frac{-25±\sqrt{625-100\left(-13320\right)}}{2\times 25}$$

Multiply -100 times -13320.

$$n=\frac{-25±\sqrt{625+1332000}}{2\times 25}$$

Add 625 to 1332000.

$$n=\frac{-25±\sqrt{1332625}}{2\times 25}$$

Take the square root of 1332625.

$$n=\frac{-25±5\sqrt{53305}}{2\times 25}$$

Multiply 2 times 25.

$$n=\frac{-25±5\sqrt{53305}}{50}$$

Now solve the equation n=\frac{-25±5\sqrt{53305}}{50} when ± is plus. Add -25 to 5\sqrt{53305}.

$$n=\frac{5\sqrt{53305}-25}{50}$$

Divide -25+5\sqrt{53305} by 50.

$$n=\frac{\sqrt{53305}}{10}-\frac{1}{2}$$

Now solve the equation n=\frac{-25±5\sqrt{53305}}{50} when ± is minus. Subtract 5\sqrt{53305} from -25.

$$n=\frac{-5\sqrt{53305}-25}{50}$$

Divide -25-5\sqrt{53305} by 50.

$$n=-\frac{\sqrt{53305}}{10}-\frac{1}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{53305}}{10}-\frac{1}{2} n=-\frac{\sqrt{53305}}{10}-\frac{1}{2}$$

Steps for Completing the Square

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

$$25n^{2}+25n=13320$$

Divide both sides by 25.

$$\frac{25n^{2}+25n}{25}=\frac{13320}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$n^{2}+\frac{25}{25}n=\frac{13320}{25}$$

Divide 25 by 25.

$$n^{2}+n=\frac{13320}{25}$$

Reduce the fraction \frac{13320}{25} to lowest terms by extracting and canceling out 5.

$$n^{2}+n=\frac{2664}{5}$$

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.

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=\frac{2664}{5}+\left(\frac{1}{2}\right)^{2}$$

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

$$n^{2}+n+\frac{1}{4}=\frac{2664}{5}+\frac{1}{4}$$

Add \frac{2664}{5} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+n+\frac{1}{4}=\frac{10661}{20}$$

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}.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{10661}{20}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{10661}{20}}$$

Simplify.

$$n+\frac{1}{2}=\frac{\sqrt{53305}}{10} n+\frac{1}{2}=-\frac{\sqrt{53305}}{10}$$

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

$$n=\frac{\sqrt{53305}}{10}-\frac{1}{2} n=-\frac{\sqrt{53305}}{10}-\frac{1}{2}$$