Solve 3m^2=-16m-21

$$3 m ^ { 2 } = - 16 m - 21$$

Solve for m

$$m = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$m=-3$$

Steps Using Factoring By Grouping

Add 16m to both sides.

$$3m^{2}+16m=-21$$

Add 21 to both sides.

$$3m^{2}+16m+21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3m^{2}+am+bm+21. To find a and b, set up a system to be solved.

$$a+b=16 ab=3\times 21=63$$

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 63.

$$1,63 3,21 7,9$$

Calculate the sum for each pair.

$$1+63=64 3+21=24 7+9=16$$

The solution is the pair that gives sum 16.

$$a=7 b=9$$

Rewrite 3m^{2}+16m+21 as \left(3m^{2}+7m\right)+\left(9m+21\right).

$$\left(3m^{2}+7m\right)+\left(9m+21\right)$$

Factor out m in the first and 3 in the second group.

$$m\left(3m+7\right)+3\left(3m+7\right)$$

Factor out common term 3m+7 by using distributive property.

$$\left(3m+7\right)\left(m+3\right)$$

To find equation solutions, solve 3m+7=0 and m+3=0.

$$m=-\frac{7}{3} m=-3$$

Steps Using the Quadratic Formula

Add 16m to both sides.

$$3m^{2}+16m=-21$$

Add 21 to both sides.

$$3m^{2}+16m+21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 16 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-16±\sqrt{16^{2}-4\times 3\times 21}}{2\times 3}$$

Square 16.

$$m=\frac{-16±\sqrt{256-4\times 3\times 21}}{2\times 3}$$

Multiply -4 times 3.

$$m=\frac{-16±\sqrt{256-12\times 21}}{2\times 3}$$

Multiply -12 times 21.

$$m=\frac{-16±\sqrt{256-252}}{2\times 3}$$

Add 256 to -252.

$$m=\frac{-16±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$m=\frac{-16±2}{2\times 3}$$

Multiply 2 times 3.

$$m=\frac{-16±2}{6}$$

Now solve the equation m=\frac{-16±2}{6} when ± is plus. Add -16 to 2.

$$m=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

$$m=-\frac{7}{3}$$

Now solve the equation m=\frac{-16±2}{6} when ± is minus. Subtract 2 from -16.

$$m=-\frac{18}{6}$$

Divide -18 by 6.

$$m=-3$$

The equation is now solved.

$$m=-\frac{7}{3} m=-3$$

Steps for Completing the Square

Add 16m to both sides.

$$3m^{2}+16m=-21$$

Divide both sides by 3.

$$\frac{3m^{2}+16m}{3}=-\frac{21}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$m^{2}+\frac{16}{3}m=-\frac{21}{3}$$

Divide -21 by 3.

$$m^{2}+\frac{16}{3}m=-7$$

Divide \frac{16}{3}, the coefficient of the x term, by 2 to get \frac{8}{3}. Then add the square of \frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+\frac{16}{3}m+\left(\frac{8}{3}\right)^{2}=-7+\left(\frac{8}{3}\right)^{2}$$

Square \frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+\frac{16}{3}m+\frac{64}{9}=-7+\frac{64}{9}$$

Add -7 to \frac{64}{9}.

$$m^{2}+\frac{16}{3}m+\frac{64}{9}=\frac{1}{9}$$

Factor m^{2}+\frac{16}{3}m+\frac{64}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m+\frac{8}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{8}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$m+\frac{8}{3}=\frac{1}{3} m+\frac{8}{3}=-\frac{1}{3}$$

Subtract \frac{8}{3} from both sides of the equation.

$$m=-\frac{7}{3} m=-3$$