Solve n^2=18n+40

$$n ^ { 2 } = 18 n + 40$$

Solve for n

$$n=-2$$

$$n=20$$

Steps Using Factoring

Subtract 18n from both sides.

$$n^{2}-18n=40$$

Subtract 40 from both sides.

$$n^{2}-18n-40=0$$

To solve the equation, factor n^{2}-18n-40 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=-18 ab=-40$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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 -18.

$$a=-20 b=2$$

Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.

$$\left(n-20\right)\left(n+2\right)$$

To find equation solutions, solve n-20=0 and n+2=0.

$$n=20 n=-2$$

Steps Using Factoring By Grouping

Subtract 18n from both sides.

$$n^{2}-18n=40$$

Subtract 40 from both sides.

$$n^{2}-18n-40=0$$

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.

$$a+b=-18 ab=1\left(-40\right)=-40$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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 -18.

$$a=-20 b=2$$

Rewrite n^{2}-18n-40 as \left(n^{2}-20n\right)+\left(2n-40\right).

$$\left(n^{2}-20n\right)+\left(2n-40\right)$$

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

$$n\left(n-20\right)+2\left(n-20\right)$$

Factor out common term n-20 by using distributive property.

$$\left(n-20\right)\left(n+2\right)$$

To find equation solutions, solve n-20=0 and n+2=0.

$$n=20 n=-2$$

Steps Using the Quadratic Formula

Subtract 18n from both sides.

$$n^{2}-18n=40$$

Subtract 40 from both sides.

$$n^{2}-18n-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-40\right)}}{2}$$

Square -18.

$$n=\frac{-\left(-18\right)±\sqrt{324-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$n=\frac{-\left(-18\right)±\sqrt{324+160}}{2}$$

Add 324 to 160.

$$n=\frac{-\left(-18\right)±\sqrt{484}}{2}$$

Take the square root of 484.

$$n=\frac{-\left(-18\right)±22}{2}$$

The opposite of -18 is 18.

$$n=\frac{18±22}{2}$$

Now solve the equation n=\frac{18±22}{2} when ± is plus. Add 18 to 22.

$$n=\frac{40}{2}$$

Divide 40 by 2.

$$n=20$$

Now solve the equation n=\frac{18±22}{2} when ± is minus. Subtract 22 from 18.

$$n=-\frac{4}{2}$$

Divide -4 by 2.

$$n=-2$$

The equation is now solved.

$$n=20 n=-2$$

Steps for Completing the Square

Subtract 18n from both sides.

$$n^{2}-18n=40$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-18n+\left(-9\right)^{2}=40+\left(-9\right)^{2}$$

Square -9.

$$n^{2}-18n+81=40+81$$

Add 40 to 81.

$$n^{2}-18n+81=121$$

Factor n^{2}-18n+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-9\right)^{2}=121$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-9\right)^{2}}=\sqrt{121}$$

Simplify.

$$n-9=11 n-9=-11$$

Add 9 to both sides of the equation.

$$n=20 n=-2$$