Solve -99=-n^2+2n

$$- 99 = - n ^ { 2 } + 2 n$$

Solve for n

$$n=-9$$

$$n=11$$

Steps Using Factoring By Grouping

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

$$-n^{2}+2n=-99$$

Add 99 to both sides.

$$-n^{2}+2n+99=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn+99. To find a and b, set up a system to be solved.

$$a+b=2 ab=-99=-99$$

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 -99.

$$-1,99 -3,33 -9,11$$

Calculate the sum for each pair.

$$-1+99=98 -3+33=30 -9+11=2$$

The solution is the pair that gives sum 2.

$$a=11 b=-9$$

Rewrite -n^{2}+2n+99 as \left(-n^{2}+11n\right)+\left(-9n+99\right).

$$\left(-n^{2}+11n\right)+\left(-9n+99\right)$$

Factor out -n in the first and -9 in the second group.

$$-n\left(n-11\right)-9\left(n-11\right)$$

Factor out common term n-11 by using distributive property.

$$\left(n-11\right)\left(-n-9\right)$$

To find equation solutions, solve n-11=0 and -n-9=0.

$$n=11 n=-9$$

Steps Using the Quadratic Formula

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

$$-n^{2}+2n=-99$$

Add 99 to both sides.

$$-n^{2}+2n+99=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 2 for b, and 99 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 99}}{2\left(-1\right)}$$

Square 2.

$$n=\frac{-2±\sqrt{4-4\left(-1\right)\times 99}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-2±\sqrt{4+4\times 99}}{2\left(-1\right)}$$

Multiply 4 times 99.

$$n=\frac{-2±\sqrt{4+396}}{2\left(-1\right)}$$

Add 4 to 396.

$$n=\frac{-2±\sqrt{400}}{2\left(-1\right)}$$

Take the square root of 400.

$$n=\frac{-2±20}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{-2±20}{-2}$$

Now solve the equation n=\frac{-2±20}{-2} when ± is plus. Add -2 to 20.

$$n=\frac{18}{-2}$$

Divide 18 by -2.

$$n=-9$$

Now solve the equation n=\frac{-2±20}{-2} when ± is minus. Subtract 20 from -2.

$$n=-\frac{22}{-2}$$

Divide -22 by -2.

$$n=11$$

The equation is now solved.

$$n=-9 n=11$$

Steps for Completing the Square

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

$$-n^{2}+2n=-99$$

Divide both sides by -1.

$$\frac{-n^{2}+2n}{-1}=-\frac{99}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\frac{2}{-1}n=-\frac{99}{-1}$$

Divide 2 by -1.

$$n^{2}-2n=-\frac{99}{-1}$$

Divide -99 by -1.

$$n^{2}-2n=99$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-2n+1=99+1$$

Add 99 to 1.

$$n^{2}-2n+1=100$$

Factor n^{2}-2n+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-1\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-1\right)^{2}}=\sqrt{100}$$

Simplify.

$$n-1=10 n-1=-10$$

Add 1 to both sides of the equation.

$$n=11 n=-9$$