Solve n^2+7n/2=270

$$\frac { n ^ { 2 } + 7 n } { 2 } = 270$$

Solve for n

$$n=-27$$

$$n=20$$

Steps Using Factoring

Multiply both sides by 2.

$$n^{2}+7n=270\times 2$$

Multiply 270 and 2 to get 540.

$$n^{2}+7n=540$$

Subtract 540 from both sides.

$$n^{2}+7n-540=0$$

To solve the equation, factor n^{2}+7n-540 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=7 ab=-540$$

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 -540.

$$-1,540 -2,270 -3,180 -4,135 -5,108 -6,90 -9,60 -10,54 -12,45 -15,36 -18,30 -20,27$$

Calculate the sum for each pair.

$$-1+540=539 -2+270=268 -3+180=177 -4+135=131 -5+108=103 -6+90=84 -9+60=51 -10+54=44 -12+45=33 -15+36=21 -18+30=12 -20+27=7$$

The solution is the pair that gives sum 7.

$$a=-20 b=27$$

Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.

$$\left(n-20\right)\left(n+27\right)$$

To find equation solutions, solve n-20=0 and n+27=0.

$$n=20 n=-27$$

Steps Using Factoring By Grouping

Multiply both sides by 2.

$$n^{2}+7n=270\times 2$$

Multiply 270 and 2 to get 540.

$$n^{2}+7n=540$$

Subtract 540 from both sides.

$$n^{2}+7n-540=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-540. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\left(-540\right)=-540$$

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 -540.

$$-1,540 -2,270 -3,180 -4,135 -5,108 -6,90 -9,60 -10,54 -12,45 -15,36 -18,30 -20,27$$

Calculate the sum for each pair.

$$-1+540=539 -2+270=268 -3+180=177 -4+135=131 -5+108=103 -6+90=84 -9+60=51 -10+54=44 -12+45=33 -15+36=21 -18+30=12 -20+27=7$$

The solution is the pair that gives sum 7.

$$a=-20 b=27$$

Rewrite n^{2}+7n-540 as \left(n^{2}-20n\right)+\left(27n-540\right).

$$\left(n^{2}-20n\right)+\left(27n-540\right)$$

Factor out n in the first and 27 in the second group.

$$n\left(n-20\right)+27\left(n-20\right)$$

Factor out common term n-20 by using distributive property.

$$\left(n-20\right)\left(n+27\right)$$

To find equation solutions, solve n-20=0 and n+27=0.

$$n=20 n=-27$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$n^{2}+7n=270\times 2$$

Multiply 270 and 2 to get 540.

$$n^{2}+7n=540$$

Subtract 540 from both sides.

$$n^{2}+7n-540=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and -540 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-7±\sqrt{7^{2}-4\left(-540\right)}}{2}$$

Square 7.

$$n=\frac{-7±\sqrt{49-4\left(-540\right)}}{2}$$

Multiply -4 times -540.

$$n=\frac{-7±\sqrt{49+2160}}{2}$$

Add 49 to 2160.

$$n=\frac{-7±\sqrt{2209}}{2}$$

Take the square root of 2209.

$$n=\frac{-7±47}{2}$$

Now solve the equation n=\frac{-7±47}{2} when ± is plus. Add -7 to 47.

$$n=\frac{40}{2}$$

Divide 40 by 2.

$$n=20$$

Now solve the equation n=\frac{-7±47}{2} when ± is minus. Subtract 47 from -7.

$$n=-\frac{54}{2}$$

Divide -54 by 2.

$$n=-27$$

The equation is now solved.

$$n=20 n=-27$$

Steps for Completing the Square

Multiply both sides by 2.

$$n^{2}+7n=270\times 2$$

Multiply 270 and 2 to get 540.

$$n^{2}+7n=540$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+7n+\left(\frac{7}{2}\right)^{2}=540+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+7n+\frac{49}{4}=540+\frac{49}{4}$$

Add 540 to \frac{49}{4}.

$$n^{2}+7n+\frac{49}{4}=\frac{2209}{4}$$

Factor n^{2}+7n+\frac{49}{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}.

$$\left(n+\frac{7}{2}\right)^{2}=\frac{2209}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{7}{2}\right)^{2}}=\sqrt{\frac{2209}{4}}$$

Simplify.

$$n+\frac{7}{2}=\frac{47}{2} n+\frac{7}{2}=-\frac{47}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$n=20 n=-27$$