Solve 792/3=n^2+n-1/3

$$79 \frac { 2 } { 3 } = \frac { n ^ { 2 } + n - 1 } { 3 }$$

Solve for n

$$n=-16$$

$$n=15$$

Steps Using Factoring

Multiply both sides of the equation by 3.

$$79\times 3+2=n^{2}+n-1$$

Multiply 79 and 3 to get 237.

$$237+2=n^{2}+n-1$$

Add 237 and 2 to get 239.

$$239=n^{2}+n-1$$

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

$$n^{2}+n-1=239$$

Subtract 239 from both sides.

$$n^{2}+n-1-239=0$$

Subtract 239 from -1 to get -240.

$$n^{2}+n-240=0$$

To solve the equation, factor n^{2}+n-240 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.

$$a+b=1 ab=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 1.

$$a=-15 b=16$$

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

$$\left(n-15\right)\left(n+16\right)$$

To find equation solutions, solve n-15=0 and n+16=0.

$$n=15 n=-16$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 3.

$$79\times 3+2=n^{2}+n-1$$

Multiply 79 and 3 to get 237.

$$237+2=n^{2}+n-1$$

Add 237 and 2 to get 239.

$$239=n^{2}+n-1$$

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

$$n^{2}+n-1=239$$

Subtract 239 from both sides.

$$n^{2}+n-1-239=0$$

Subtract 239 from -1 to get -240.

$$n^{2}+n-240=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-240. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-240\right)=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 1.

$$a=-15 b=16$$

Rewrite n^{2}+n-240 as \left(n^{2}-15n\right)+\left(16n-240\right).

$$\left(n^{2}-15n\right)+\left(16n-240\right)$$

Factor out n in the first and 16 in the second group.

$$n\left(n-15\right)+16\left(n-15\right)$$

Factor out common term n-15 by using distributive property.

$$\left(n-15\right)\left(n+16\right)$$

To find equation solutions, solve n-15=0 and n+16=0.

$$n=15 n=-16$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$79\times 3+2=n^{2}+n-1$$

Multiply 79 and 3 to get 237.

$$237+2=n^{2}+n-1$$

Add 237 and 2 to get 239.

$$239=n^{2}+n-1$$

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

$$n^{2}+n-1=239$$

Subtract 239 from both sides.

$$n^{2}+n-1-239=0$$

Subtract 239 from -1 to get -240.

$$n^{2}+n-240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-1±\sqrt{1^{2}-4\left(-240\right)}}{2}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\left(-240\right)}}{2}$$

Multiply -4 times -240.

$$n=\frac{-1±\sqrt{1+960}}{2}$$

Add 1 to 960.

$$n=\frac{-1±\sqrt{961}}{2}$$

Take the square root of 961.

$$n=\frac{-1±31}{2}$$

Now solve the equation n=\frac{-1±31}{2} when ± is plus. Add -1 to 31.

$$n=\frac{30}{2}$$

Divide 30 by 2.

$$n=15$$

Now solve the equation n=\frac{-1±31}{2} when ± is minus. Subtract 31 from -1.

$$n=-\frac{32}{2}$$

Divide -32 by 2.

$$n=-16$$

The equation is now solved.

$$n=15 n=-16$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$79\times 3+2=n^{2}+n-1$$

Multiply 79 and 3 to get 237.

$$237+2=n^{2}+n-1$$

Add 237 and 2 to get 239.

$$239=n^{2}+n-1$$

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

$$n^{2}+n-1=239$$

Add 1 to both sides.

$$n^{2}+n=239+1$$

Add 239 and 1 to get 240.

$$n^{2}+n=240$$

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.

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=240+\left(\frac{1}{2}\right)^{2}$$

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

$$n^{2}+n+\frac{1}{4}=240+\frac{1}{4}$$

Add 240 to \frac{1}{4}.

$$n^{2}+n+\frac{1}{4}=\frac{961}{4}$$

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}.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{961}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{961}{4}}$$

Simplify.

$$n+\frac{1}{2}=\frac{31}{2} n+\frac{1}{2}=-\frac{31}{2}$$

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

$$n=15 n=-16$$