Solve 2n^2+67n-36n-1206=0

$$2 n ^ { 2 } + 67 n - 36 n - 1206 = 0$$

Solve for n

$$n = -\frac{67}{2} = -33\frac{1}{2} = -33.5$$

$$n=18$$

Steps Using Factoring By Grouping

Combine 67n and -36n to get 31n.

$$2n^{2}+31n-1206=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2n^{2}+an+bn-1206. To find a and b, set up a system to be solved.

$$a+b=31 ab=2\left(-1206\right)=-2412$$

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 -2412.

$$-1,2412 -2,1206 -3,804 -4,603 -6,402 -9,268 -12,201 -18,134 -36,67$$

Calculate the sum for each pair.

$$-1+2412=2411 -2+1206=1204 -3+804=801 -4+603=599 -6+402=396 -9+268=259 -12+201=189 -18+134=116 -36+67=31$$

The solution is the pair that gives sum 31.

$$a=-36 b=67$$

Rewrite 2n^{2}+31n-1206 as \left(2n^{2}-36n\right)+\left(67n-1206\right).

$$\left(2n^{2}-36n\right)+\left(67n-1206\right)$$

Factor out 2n in the first and 67 in the second group.

$$2n\left(n-18\right)+67\left(n-18\right)$$

Factor out common term n-18 by using distributive property.

$$\left(n-18\right)\left(2n+67\right)$$

To find equation solutions, solve n-18=0 and 2n+67=0.

$$n=18 n=-\frac{67}{2}$$

Steps Using the Quadratic Formula

Combine 67n and -36n to get 31n.

$$2n^{2}+31n-1206=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 31 for b, and -1206 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-31±\sqrt{31^{2}-4\times 2\left(-1206\right)}}{2\times 2}$$

Square 31.

$$n=\frac{-31±\sqrt{961-4\times 2\left(-1206\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-31±\sqrt{961-8\left(-1206\right)}}{2\times 2}$$

Multiply -8 times -1206.

$$n=\frac{-31±\sqrt{961+9648}}{2\times 2}$$

Add 961 to 9648.

$$n=\frac{-31±\sqrt{10609}}{2\times 2}$$

Take the square root of 10609.

$$n=\frac{-31±103}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{-31±103}{4}$$

Now solve the equation n=\frac{-31±103}{4} when ± is plus. Add -31 to 103.

$$n=\frac{72}{4}$$

Divide 72 by 4.

$$n=18$$

Now solve the equation n=\frac{-31±103}{4} when ± is minus. Subtract 103 from -31.

$$n=-\frac{134}{4}$$

Reduce the fraction \frac{-134}{4} to lowest terms by extracting and canceling out 2.

$$n=-\frac{67}{2}$$

The equation is now solved.

$$n=18 n=-\frac{67}{2}$$

Steps for Completing the Square

Combine 67n and -36n to get 31n.

$$2n^{2}+31n-1206=0$$

Add 1206 to both sides. Anything plus zero gives itself.

$$2n^{2}+31n=1206$$

Divide both sides by 2.

$$\frac{2n^{2}+31n}{2}=\frac{1206}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}+\frac{31}{2}n=\frac{1206}{2}$$

Divide 1206 by 2.

$$n^{2}+\frac{31}{2}n=603$$

Divide \frac{31}{2}, the coefficient of the x term, by 2 to get \frac{31}{4}. Then add the square of \frac{31}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{31}{2}n+\left(\frac{31}{4}\right)^{2}=603+\left(\frac{31}{4}\right)^{2}$$

Square \frac{31}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{31}{2}n+\frac{961}{16}=603+\frac{961}{16}$$

Add 603 to \frac{961}{16}.

$$n^{2}+\frac{31}{2}n+\frac{961}{16}=\frac{10609}{16}$$

Factor n^{2}+\frac{31}{2}n+\frac{961}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{31}{4}\right)^{2}=\frac{10609}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{31}{4}\right)^{2}}=\sqrt{\frac{10609}{16}}$$

Simplify.

$$n+\frac{31}{4}=\frac{103}{4} n+\frac{31}{4}=-\frac{103}{4}$$

Subtract \frac{31}{4} from both sides of the equation.

$$n=18 n=-\frac{67}{2}$$