Solve 19m^2+5m-12=-6m^2

$$19 m ^ { 2 } + 5 m - 12 = - 6 m ^ { 2 }$$

Solve for m

$$m=-\frac{4}{5}=-0.8$$

$$m=\frac{3}{5}=0.6$$

Steps Using Factoring By Grouping

Add 6m^{2} to both sides.

$$19m^{2}+5m-12+6m^{2}=0$$

Combine 19m^{2} and 6m^{2} to get 25m^{2}.

$$25m^{2}+5m-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25m^{2}+am+bm-12. To find a and b, set up a system to be solved.

$$a+b=5 ab=25\left(-12\right)=-300$$

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 -300.

$$-1,300 -2,150 -3,100 -4,75 -5,60 -6,50 -10,30 -12,25 -15,20$$

Calculate the sum for each pair.

$$-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5$$

The solution is the pair that gives sum 5.

$$a=-15 b=20$$

Rewrite 25m^{2}+5m-12 as \left(25m^{2}-15m\right)+\left(20m-12\right).

$$\left(25m^{2}-15m\right)+\left(20m-12\right)$$

Factor out 5m in the first and 4 in the second group.

$$5m\left(5m-3\right)+4\left(5m-3\right)$$

Factor out common term 5m-3 by using distributive property.

$$\left(5m-3\right)\left(5m+4\right)$$

To find equation solutions, solve 5m-3=0 and 5m+4=0.

$$m=\frac{3}{5} m=-\frac{4}{5}$$

Steps Using the Quadratic Formula

Add 6m^{2} to both sides.

$$19m^{2}+5m-12+6m^{2}=0$$

Combine 19m^{2} and 6m^{2} to get 25m^{2}.

$$25m^{2}+5m-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 5 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-5±\sqrt{5^{2}-4\times 25\left(-12\right)}}{2\times 25}$$

Square 5.

$$m=\frac{-5±\sqrt{25-4\times 25\left(-12\right)}}{2\times 25}$$

Multiply -4 times 25.

$$m=\frac{-5±\sqrt{25-100\left(-12\right)}}{2\times 25}$$

Multiply -100 times -12.

$$m=\frac{-5±\sqrt{25+1200}}{2\times 25}$$

Add 25 to 1200.

$$m=\frac{-5±\sqrt{1225}}{2\times 25}$$

Take the square root of 1225.

$$m=\frac{-5±35}{2\times 25}$$

Multiply 2 times 25.

$$m=\frac{-5±35}{50}$$

Now solve the equation m=\frac{-5±35}{50} when ± is plus. Add -5 to 35.

$$m=\frac{30}{50}$$

Reduce the fraction \frac{30}{50} to lowest terms by extracting and canceling out 10.

$$m=\frac{3}{5}$$

Now solve the equation m=\frac{-5±35}{50} when ± is minus. Subtract 35 from -5.

$$m=-\frac{40}{50}$$

Reduce the fraction \frac{-40}{50} to lowest terms by extracting and canceling out 10.

$$m=-\frac{4}{5}$$

The equation is now solved.

$$m=\frac{3}{5} m=-\frac{4}{5}$$

Steps for Completing the Square

Add 6m^{2} to both sides.

$$19m^{2}+5m-12+6m^{2}=0$$

Combine 19m^{2} and 6m^{2} to get 25m^{2}.

$$25m^{2}+5m-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$25m^{2}+5m=12$$

Divide both sides by 25.

$$\frac{25m^{2}+5m}{25}=\frac{12}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$m^{2}+\frac{5}{25}m=\frac{12}{25}$$

Reduce the fraction \frac{5}{25} to lowest terms by extracting and canceling out 5.

$$m^{2}+\frac{1}{5}m=\frac{12}{25}$$

Divide \frac{1}{5}, the coefficient of the x term, by 2 to get \frac{1}{10}. Then add the square of \frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+\frac{1}{5}m+\left(\frac{1}{10}\right)^{2}=\frac{12}{25}+\left(\frac{1}{10}\right)^{2}$$

Square \frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+\frac{1}{5}m+\frac{1}{100}=\frac{12}{25}+\frac{1}{100}$$

Add \frac{12}{25} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}+\frac{1}{5}m+\frac{1}{100}=\frac{49}{100}$$

Factor m^{2}+\frac{1}{5}m+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m+\frac{1}{10}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{1}{10}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$m+\frac{1}{10}=\frac{7}{10} m+\frac{1}{10}=-\frac{7}{10}$$

Subtract \frac{1}{10} from both sides of the equation.

$$m=\frac{3}{5} m=-\frac{4}{5}$$