Solve 116=53n/2-3n^2/2

$$116 = \frac { 53 n } { 2 } - \frac { 3 n ^ { 2 } } { 2 }$$

Solve for n

$$n=8$$

$$n = \frac{29}{3} = 9\frac{2}{3} \approx 9.666666667$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

$$232=53n-3n^{2}$$

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

$$53n-3n^{2}=232$$

Subtract 232 from both sides.

$$53n-3n^{2}-232=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3n^{2}+53n-232=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3n^{2}+an+bn-232. To find a and b, set up a system to be solved.

$$a+b=53 ab=-3\left(-232\right)=696$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 696.

$$1,696 2,348 3,232 4,174 6,116 8,87 12,58 24,29$$

Calculate the sum for each pair.

$$1+696=697 2+348=350 3+232=235 4+174=178 6+116=122 8+87=95 12+58=70 24+29=53$$

The solution is the pair that gives sum 53.

$$a=29 b=24$$

Rewrite -3n^{2}+53n-232 as \left(-3n^{2}+29n\right)+\left(24n-232\right).

$$\left(-3n^{2}+29n\right)+\left(24n-232\right)$$

Factor out -n in the first and 8 in the second group.

$$-n\left(3n-29\right)+8\left(3n-29\right)$$

Factor out common term 3n-29 by using distributive property.

$$\left(3n-29\right)\left(-n+8\right)$$

To find equation solutions, solve 3n-29=0 and -n+8=0.

$$n=\frac{29}{3} n=8$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$232=53n-3n^{2}$$

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

$$53n-3n^{2}=232$$

Subtract 232 from both sides.

$$53n-3n^{2}-232=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3n^{2}+53n-232=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 53 for b, and -232 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-53±\sqrt{53^{2}-4\left(-3\right)\left(-232\right)}}{2\left(-3\right)}$$

Square 53.

$$n=\frac{-53±\sqrt{2809-4\left(-3\right)\left(-232\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$n=\frac{-53±\sqrt{2809+12\left(-232\right)}}{2\left(-3\right)}$$

Multiply 12 times -232.

$$n=\frac{-53±\sqrt{2809-2784}}{2\left(-3\right)}$$

Add 2809 to -2784.

$$n=\frac{-53±\sqrt{25}}{2\left(-3\right)}$$

Take the square root of 25.

$$n=\frac{-53±5}{2\left(-3\right)}$$

Multiply 2 times -3.

$$n=\frac{-53±5}{-6}$$

Now solve the equation n=\frac{-53±5}{-6} when ± is plus. Add -53 to 5.

$$n=-\frac{48}{-6}$$

Divide -48 by -6.

$$n=8$$

Now solve the equation n=\frac{-53±5}{-6} when ± is minus. Subtract 5 from -53.

$$n=-\frac{58}{-6}$$

Reduce the fraction \frac{-58}{-6} to lowest terms by extracting and canceling out 2.

$$n=\frac{29}{3}$$

The equation is now solved.

$$n=8 n=\frac{29}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$232=53n-3n^{2}$$

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

$$53n-3n^{2}=232$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-3n^{2}+53n=232$$

Divide both sides by -3.

$$\frac{-3n^{2}+53n}{-3}=\frac{232}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$n^{2}+\frac{53}{-3}n=\frac{232}{-3}$$

Divide 53 by -3.

$$n^{2}-\frac{53}{3}n=\frac{232}{-3}$$

Divide 232 by -3.

$$n^{2}-\frac{53}{3}n=-\frac{232}{3}$$

Divide -\frac{53}{3}, the coefficient of the x term, by 2 to get -\frac{53}{6}. Then add the square of -\frac{53}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{53}{3}n+\left(-\frac{53}{6}\right)^{2}=-\frac{232}{3}+\left(-\frac{53}{6}\right)^{2}$$

Square -\frac{53}{6} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{53}{3}n+\frac{2809}{36}=-\frac{232}{3}+\frac{2809}{36}$$

Add -\frac{232}{3} to \frac{2809}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{53}{3}n+\frac{2809}{36}=\frac{25}{36}$$

Factor n^{2}-\frac{53}{3}n+\frac{2809}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{53}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{53}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$n-\frac{53}{6}=\frac{5}{6} n-\frac{53}{6}=-\frac{5}{6}$$

Add \frac{53}{6} to both sides of the equation.

$$n=\frac{29}{3} n=8$$