Solve 40+3n=n^2

$$40 + 3 n = n ^ { 2 }$$

Solve for n

$$n=-5$$

$$n=8$$

Steps Using Factoring By Grouping

Subtract n^{2} from both sides.

$$40+3n-n^{2}=0$$

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

$$-n^{2}+3n+40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn+40. To find a and b, set up a system to be solved.

$$a+b=3 ab=-40=-40$$

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 -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=8 b=-5$$

Rewrite -n^{2}+3n+40 as \left(-n^{2}+8n\right)+\left(-5n+40\right).

$$\left(-n^{2}+8n\right)+\left(-5n+40\right)$$

Factor out -n in the first and -5 in the second group.

$$-n\left(n-8\right)-5\left(n-8\right)$$

Factor out common term n-8 by using distributive property.

$$\left(n-8\right)\left(-n-5\right)$$

To find equation solutions, solve n-8=0 and -n-5=0.

$$n=8 n=-5$$

Steps Using the Quadratic Formula

Subtract n^{2} from both sides.

$$40+3n-n^{2}=0$$

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.

$$-n^{2}+3n+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\times 40}}{2\left(-1\right)}$$

Square 3.

$$n=\frac{-3±\sqrt{9-4\left(-1\right)\times 40}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-3±\sqrt{9+4\times 40}}{2\left(-1\right)}$$

Multiply 4 times 40.

$$n=\frac{-3±\sqrt{9+160}}{2\left(-1\right)}$$

Add 9 to 160.

$$n=\frac{-3±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$n=\frac{-3±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{-3±13}{-2}$$

Now solve the equation n=\frac{-3±13}{-2} when ± is plus. Add -3 to 13.

$$n=\frac{10}{-2}$$

Divide 10 by -2.

$$n=-5$$

Now solve the equation n=\frac{-3±13}{-2} when ± is minus. Subtract 13 from -3.

$$n=-\frac{16}{-2}$$

Divide -16 by -2.

$$n=8$$

The equation is now solved.

$$n=-5 n=8$$

Steps for Completing the Square

Subtract n^{2} from both sides.

$$40+3n-n^{2}=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$3n-n^{2}=-40$$

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.

$$-n^{2}+3n=-40$$

Divide both sides by -1.

$$\frac{-n^{2}+3n}{-1}=-\frac{40}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\frac{3}{-1}n=-\frac{40}{-1}$$

Divide 3 by -1.

$$n^{2}-3n=-\frac{40}{-1}$$

Divide -40 by -1.

$$n^{2}-3n=40$$

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

$$n^{2}-3n+\left(-\frac{3}{2}\right)^{2}=40+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-3n+\frac{9}{4}=40+\frac{9}{4}$$

Add 40 to \frac{9}{4}.

$$n^{2}-3n+\frac{9}{4}=\frac{169}{4}$$

Factor n^{2}-3n+\frac{9}{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}.

$$\left(n-\frac{3}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$n-\frac{3}{2}=\frac{13}{2} n-\frac{3}{2}=-\frac{13}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$n=8 n=-5$$