Solve -27=2m^2+21m

$$- 27 = 2 m ^ { 2 } + 21 m$$

Solve for m

$$m = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$m=-9$$

Steps Using Factoring By Grouping

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

$$2m^{2}+21m=-27$$

Add 27 to both sides.

$$2m^{2}+21m+27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2m^{2}+am+bm+27. To find a and b, set up a system to be solved.

$$a+b=21 ab=2\times 27=54$$

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 54.

$$1,54 2,27 3,18 6,9$$

Calculate the sum for each pair.

$$1+54=55 2+27=29 3+18=21 6+9=15$$

The solution is the pair that gives sum 21.

$$a=3 b=18$$

Rewrite 2m^{2}+21m+27 as \left(2m^{2}+3m\right)+\left(18m+27\right).

$$\left(2m^{2}+3m\right)+\left(18m+27\right)$$

Factor out m in the first and 9 in the second group.

$$m\left(2m+3\right)+9\left(2m+3\right)$$

Factor out common term 2m+3 by using distributive property.

$$\left(2m+3\right)\left(m+9\right)$$

To find equation solutions, solve 2m+3=0 and m+9=0.

$$m=-\frac{3}{2} m=-9$$

Steps Using the Quadratic Formula

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

$$2m^{2}+21m=-27$$

Add 27 to both sides.

$$2m^{2}+21m+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 21 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-21±\sqrt{21^{2}-4\times 2\times 27}}{2\times 2}$$

Square 21.

$$m=\frac{-21±\sqrt{441-4\times 2\times 27}}{2\times 2}$$

Multiply -4 times 2.

$$m=\frac{-21±\sqrt{441-8\times 27}}{2\times 2}$$

Multiply -8 times 27.

$$m=\frac{-21±\sqrt{441-216}}{2\times 2}$$

Add 441 to -216.

$$m=\frac{-21±\sqrt{225}}{2\times 2}$$

Take the square root of 225.

$$m=\frac{-21±15}{2\times 2}$$

Multiply 2 times 2.

$$m=\frac{-21±15}{4}$$

Now solve the equation m=\frac{-21±15}{4} when ± is plus. Add -21 to 15.

$$m=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$m=-\frac{3}{2}$$

Now solve the equation m=\frac{-21±15}{4} when ± is minus. Subtract 15 from -21.

$$m=-\frac{36}{4}$$

Divide -36 by 4.

$$m=-9$$

The equation is now solved.

$$m=-\frac{3}{2} m=-9$$

Steps for Completing the Square

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

$$2m^{2}+21m=-27$$

Divide both sides by 2.

$$\frac{2m^{2}+21m}{2}=-\frac{27}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$m^{2}+\frac{21}{2}m=-\frac{27}{2}$$

Divide \frac{21}{2}, the coefficient of the x term, by 2 to get \frac{21}{4}. Then add the square of \frac{21}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+\frac{21}{2}m+\left(\frac{21}{4}\right)^{2}=-\frac{27}{2}+\left(\frac{21}{4}\right)^{2}$$

Square \frac{21}{4} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+\frac{21}{2}m+\frac{441}{16}=-\frac{27}{2}+\frac{441}{16}$$

Add -\frac{27}{2} to \frac{441}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}+\frac{21}{2}m+\frac{441}{16}=\frac{225}{16}$$

Factor m^{2}+\frac{21}{2}m+\frac{441}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m+\frac{21}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{21}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$m+\frac{21}{4}=\frac{15}{4} m+\frac{21}{4}=-\frac{15}{4}$$

Subtract \frac{21}{4} from both sides of the equation.

$$m=-\frac{3}{2} m=-9$$