Solve 0=5n^2+1205n-90300

$$0 = 5 n ^ { 2 } + 1205 n - 90300$$

Solve for n

$$n=-301$$

$$n=60$$

Steps Using Factoring By Grouping

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

$$5n^{2}+1205n-90300=0$$

Divide both sides by 5.

$$n^{2}+241n-18060=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-18060. To find a and b, set up a system to be solved.

$$a+b=241 ab=1\left(-18060\right)=-18060$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18060.

$$-1,18060 -2,9030 -3,6020 -4,4515 -5,3612 -6,3010 -7,2580 -10,1806 -12,1505 -14,1290 -15,1204 -20,903 -21,860 -28,645 -30,602 -35,516 -42,430 -43,420 -60,301 -70,258 -84,215 -86,210 -105,172 -129,140$$

Calculate the sum for each pair.

$$-1+18060=18059 -2+9030=9028 -3+6020=6017 -4+4515=4511 -5+3612=3607 -6+3010=3004 -7+2580=2573 -10+1806=1796 -12+1505=1493 -14+1290=1276 -15+1204=1189 -20+903=883 -21+860=839 -28+645=617 -30+602=572 -35+516=481 -42+430=388 -43+420=377 -60+301=241 -70+258=188 -84+215=131 -86+210=124 -105+172=67 -129+140=11$$

The solution is the pair that gives sum 241.

$$a=-60 b=301$$

Rewrite n^{2}+241n-18060 as \left(n^{2}-60n\right)+\left(301n-18060\right).

$$\left(n^{2}-60n\right)+\left(301n-18060\right)$$

Factor out n in the first and 301 in the second group.

$$n\left(n-60\right)+301\left(n-60\right)$$

Factor out common term n-60 by using distributive property.

$$\left(n-60\right)\left(n+301\right)$$

To find equation solutions, solve n-60=0 and n+301=0.

$$n=60 n=-301$$

Steps Using the Quadratic Formula

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

$$5n^{2}+1205n-90300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 1205 for b, and -90300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-1205±\sqrt{1205^{2}-4\times 5\left(-90300\right)}}{2\times 5}$$

Square 1205.

$$n=\frac{-1205±\sqrt{1452025-4\times 5\left(-90300\right)}}{2\times 5}$$

Multiply -4 times 5.

$$n=\frac{-1205±\sqrt{1452025-20\left(-90300\right)}}{2\times 5}$$

Multiply -20 times -90300.

$$n=\frac{-1205±\sqrt{1452025+1806000}}{2\times 5}$$

Add 1452025 to 1806000.

$$n=\frac{-1205±\sqrt{3258025}}{2\times 5}$$

Take the square root of 3258025.

$$n=\frac{-1205±1805}{2\times 5}$$

Multiply 2 times 5.

$$n=\frac{-1205±1805}{10}$$

Now solve the equation n=\frac{-1205±1805}{10} when ± is plus. Add -1205 to 1805.

$$n=\frac{600}{10}$$

Divide 600 by 10.

$$n=60$$

Now solve the equation n=\frac{-1205±1805}{10} when ± is minus. Subtract 1805 from -1205.

$$n=-\frac{3010}{10}$$

Divide -3010 by 10.

$$n=-301$$

The equation is now solved.

$$n=60 n=-301$$

Steps for Completing the Square

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

$$5n^{2}+1205n-90300=0$$

Add 90300 to both sides. Anything plus zero gives itself.

$$5n^{2}+1205n=90300$$

Divide both sides by 5.

$$\frac{5n^{2}+1205n}{5}=\frac{90300}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$n^{2}+\frac{1205}{5}n=\frac{90300}{5}$$

Divide 1205 by 5.

$$n^{2}+241n=\frac{90300}{5}$$

Divide 90300 by 5.

$$n^{2}+241n=18060$$

Divide 241, the coefficient of the x term, by 2 to get \frac{241}{2}. Then add the square of \frac{241}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+241n+\left(\frac{241}{2}\right)^{2}=18060+\left(\frac{241}{2}\right)^{2}$$

Square \frac{241}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+241n+\frac{58081}{4}=18060+\frac{58081}{4}$$

Add 18060 to \frac{58081}{4}.

$$n^{2}+241n+\frac{58081}{4}=\frac{130321}{4}$$

Factor n^{2}+241n+\frac{58081}{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}.

$$\left(n+\frac{241}{2}\right)^{2}=\frac{130321}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{241}{2}\right)^{2}}=\sqrt{\frac{130321}{4}}$$

Simplify.

$$n+\frac{241}{2}=\frac{361}{2} n+\frac{241}{2}=-\frac{361}{2}$$

Subtract \frac{241}{2} from both sides of the equation.

$$n=60 n=-301$$