Solve 156/3=17n-n^2

$$\frac { 156 } { 3 } = 17 n - n ^ { 2 }$$

Solve for n

$$n=4$$

$$n=13$$

Steps Using Factoring By Grouping

Divide 156 by 3 to get 52.

$$52=17n-n^{2}$$

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

$$17n-n^{2}=52$$

Subtract 52 from both sides.

$$17n-n^{2}-52=0$$

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

$$-n^{2}+17n-52=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn-52. To find a and b, set up a system to be solved.

$$a+b=17 ab=-\left(-52\right)=52$$

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 52.

$$1,52 2,26 4,13$$

Calculate the sum for each pair.

$$1+52=53 2+26=28 4+13=17$$

The solution is the pair that gives sum 17.

$$a=13 b=4$$

Rewrite -n^{2}+17n-52 as \left(-n^{2}+13n\right)+\left(4n-52\right).

$$\left(-n^{2}+13n\right)+\left(4n-52\right)$$

Factor out -n in the first and 4 in the second group.

$$-n\left(n-13\right)+4\left(n-13\right)$$

Factor out common term n-13 by using distributive property.

$$\left(n-13\right)\left(-n+4\right)$$

To find equation solutions, solve n-13=0 and -n+4=0.

$$n=13 n=4$$

Steps Using the Quadratic Formula

Divide 156 by 3 to get 52.

$$52=17n-n^{2}$$

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

$$17n-n^{2}=52$$

Subtract 52 from both sides.

$$17n-n^{2}-52=0$$

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.

$$-n^{2}+17n-52=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 17 for b, and -52 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-17±\sqrt{17^{2}-4\left(-1\right)\left(-52\right)}}{2\left(-1\right)}$$

Square 17.

$$n=\frac{-17±\sqrt{289-4\left(-1\right)\left(-52\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-17±\sqrt{289+4\left(-52\right)}}{2\left(-1\right)}$$

Multiply 4 times -52.

$$n=\frac{-17±\sqrt{289-208}}{2\left(-1\right)}$$

Add 289 to -208.

$$n=\frac{-17±\sqrt{81}}{2\left(-1\right)}$$

Take the square root of 81.

$$n=\frac{-17±9}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{-17±9}{-2}$$

Now solve the equation n=\frac{-17±9}{-2} when ± is plus. Add -17 to 9.

$$n=-\frac{8}{-2}$$

Divide -8 by -2.

$$n=4$$

Now solve the equation n=\frac{-17±9}{-2} when ± is minus. Subtract 9 from -17.

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

Divide -26 by -2.

$$n=13$$

The equation is now solved.

$$n=4 n=13$$

Steps for Completing the Square

Divide 156 by 3 to get 52.

$$52=17n-n^{2}$$

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

$$17n-n^{2}=52$$

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.

$$-n^{2}+17n=52$$

Divide both sides by -1.

$$\frac{-n^{2}+17n}{-1}=\frac{52}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\frac{17}{-1}n=\frac{52}{-1}$$

Divide 17 by -1.

$$n^{2}-17n=\frac{52}{-1}$$

Divide 52 by -1.

$$n^{2}-17n=-52$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-17n+\left(-\frac{17}{2}\right)^{2}=-52+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-17n+\frac{289}{4}=-52+\frac{289}{4}$$

Add -52 to \frac{289}{4}.

$$n^{2}-17n+\frac{289}{4}=\frac{81}{4}$$

Factor n^{2}-17n+\frac{289}{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}.

$$\left(n-\frac{17}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{17}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$n-\frac{17}{2}=\frac{9}{2} n-\frac{17}{2}=-\frac{9}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$n=13 n=4$$