Solve 636=5n+4n^2

$$636 = 5 n + 4 n ^ { 2 }$$

Solve for n

$$n = -\frac{53}{4} = -13\frac{1}{4} = -13.25$$

$$n=12$$

Steps Using Factoring By Grouping

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

$$5n+4n^{2}=636$$

Subtract 636 from both sides.

$$5n+4n^{2}-636=0$$

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

$$4n^{2}+5n-636=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4n^{2}+an+bn-636. To find a and b, set up a system to be solved.

$$a+b=5 ab=4\left(-636\right)=-2544$$

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 -2544.

$$-1,2544 -2,1272 -3,848 -4,636 -6,424 -8,318 -12,212 -16,159 -24,106 -48,53$$

Calculate the sum for each pair.

$$-1+2544=2543 -2+1272=1270 -3+848=845 -4+636=632 -6+424=418 -8+318=310 -12+212=200 -16+159=143 -24+106=82 -48+53=5$$

The solution is the pair that gives sum 5.

$$a=-48 b=53$$

Rewrite 4n^{2}+5n-636 as \left(4n^{2}-48n\right)+\left(53n-636\right).

$$\left(4n^{2}-48n\right)+\left(53n-636\right)$$

Factor out 4n in the first and 53 in the second group.

$$4n\left(n-12\right)+53\left(n-12\right)$$

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

$$\left(n-12\right)\left(4n+53\right)$$

To find equation solutions, solve n-12=0 and 4n+53=0.

$$n=12 n=-\frac{53}{4}$$

Steps Using the Quadratic Formula

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

$$5n+4n^{2}=636$$

Subtract 636 from both sides.

$$5n+4n^{2}-636=0$$

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.

$$4n^{2}+5n-636=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 5 for b, and -636 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-5±\sqrt{5^{2}-4\times 4\left(-636\right)}}{2\times 4}$$

Square 5.

$$n=\frac{-5±\sqrt{25-4\times 4\left(-636\right)}}{2\times 4}$$

Multiply -4 times 4.

$$n=\frac{-5±\sqrt{25-16\left(-636\right)}}{2\times 4}$$

Multiply -16 times -636.

$$n=\frac{-5±\sqrt{25+10176}}{2\times 4}$$

Add 25 to 10176.

$$n=\frac{-5±\sqrt{10201}}{2\times 4}$$

Take the square root of 10201.

$$n=\frac{-5±101}{2\times 4}$$

Multiply 2 times 4.

$$n=\frac{-5±101}{8}$$

Now solve the equation n=\frac{-5±101}{8} when ± is plus. Add -5 to 101.

$$n=\frac{96}{8}$$

Divide 96 by 8.

$$n=12$$

Now solve the equation n=\frac{-5±101}{8} when ± is minus. Subtract 101 from -5.

$$n=-\frac{106}{8}$$

Reduce the fraction \frac{-106}{8} to lowest terms by extracting and canceling out 2.

$$n=-\frac{53}{4}$$

The equation is now solved.

$$n=12 n=-\frac{53}{4}$$

Steps for Completing the Square

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

$$5n+4n^{2}=636$$

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.

$$4n^{2}+5n=636$$

Divide both sides by 4.

$$\frac{4n^{2}+5n}{4}=\frac{636}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$n^{2}+\frac{5}{4}n=\frac{636}{4}$$

Divide 636 by 4.

$$n^{2}+\frac{5}{4}n=159$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{5}{4}n+\left(\frac{5}{8}\right)^{2}=159+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{5}{4}n+\frac{25}{64}=159+\frac{25}{64}$$

Add 159 to \frac{25}{64}.

$$n^{2}+\frac{5}{4}n+\frac{25}{64}=\frac{10201}{64}$$

Factor n^{2}+\frac{5}{4}n+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{5}{8}\right)^{2}=\frac{10201}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{5}{8}\right)^{2}}=\sqrt{\frac{10201}{64}}$$

Simplify.

$$n+\frac{5}{8}=\frac{101}{8} n+\frac{5}{8}=-\frac{101}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$n=12 n=-\frac{53}{4}$$