Solve 156=n^2+n

$$156 = { n }^{ 2 } +n$$

Solve for n

$$n=-13$$

$$n=12$$

Steps Using Factoring

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

$$n^{2}+n=156$$

Subtract 156 from both sides.

$$n^{2}+n-156=0$$

To solve the equation, factor n^{2}+n-156 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.

$$a+b=1 ab=-156$$

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 -156.

$$-1,156 -2,78 -3,52 -4,39 -6,26 -12,13$$

Calculate the sum for each pair.

$$-1+156=155 -2+78=76 -3+52=49 -4+39=35 -6+26=20 -12+13=1$$

The solution is the pair that gives sum 1.

$$a=-12 b=13$$

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

$$\left(n-12\right)\left(n+13\right)$$

To find equation solutions, solve n-12=0 and n+13=0.

$$n=12 n=-13$$

Steps Using Factoring By Grouping

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

$$n^{2}+n=156$$

Subtract 156 from both sides.

$$n^{2}+n-156=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-156. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-156\right)=-156$$

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 -156.

$$-1,156 -2,78 -3,52 -4,39 -6,26 -12,13$$

Calculate the sum for each pair.

$$-1+156=155 -2+78=76 -3+52=49 -4+39=35 -6+26=20 -12+13=1$$

The solution is the pair that gives sum 1.

$$a=-12 b=13$$

Rewrite n^{2}+n-156 as \left(n^{2}-12n\right)+\left(13n-156\right).

$$\left(n^{2}-12n\right)+\left(13n-156\right)$$

Factor out n in the first and 13 in the second group.

$$n\left(n-12\right)+13\left(n-12\right)$$

Factor out common term n-12 by using distributive property.

$$\left(n-12\right)\left(n+13\right)$$

To find equation solutions, solve n-12=0 and n+13=0.

$$n=12 n=-13$$

Steps Using the Quadratic Formula

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

$$n^{2}+n=156$$

Subtract 156 from both sides.

$$n^{2}+n-156=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -156 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-1±\sqrt{1^{2}-4\left(-156\right)}}{2}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\left(-156\right)}}{2}$$

Multiply -4 times -156.

$$n=\frac{-1±\sqrt{1+624}}{2}$$

Add 1 to 624.

$$n=\frac{-1±\sqrt{625}}{2}$$

Take the square root of 625.

$$n=\frac{-1±25}{2}$$

Now solve the equation n=\frac{-1±25}{2} when ± is plus. Add -1 to 25.

$$n=\frac{24}{2}$$

Divide 24 by 2.

$$n=12$$

Now solve the equation n=\frac{-1±25}{2} when ± is minus. Subtract 25 from -1.

$$n=-\frac{26}{2}$$

Divide -26 by 2.

$$n=-13$$

The equation is now solved.

$$n=12 n=-13$$

Steps for Completing the Square

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

$$n^{2}+n=156$$

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.

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=156+\left(\frac{1}{2}\right)^{2}$$

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

$$n^{2}+n+\frac{1}{4}=156+\frac{1}{4}$$

Add 156 to \frac{1}{4}.

$$n^{2}+n+\frac{1}{4}=\frac{625}{4}$$

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}.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{625}{4}}$$

Simplify.

$$n+\frac{1}{2}=\frac{25}{2} n+\frac{1}{2}=-\frac{25}{2}$$

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

$$n=12 n=-13$$