Solve 5r^2=11r+12

$$5 r ^ { 2 } = 11 r + 12$$

Solve for r

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

$$r=3$$

Steps Using Factoring By Grouping

Subtract 11r from both sides.

$$5r^{2}-11r=12$$

Subtract 12 from both sides.

$$5r^{2}-11r-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5r^{2}+ar+br-12. To find a and b, set up a system to be solved.

$$a+b=-11 ab=5\left(-12\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -11.

$$a=-15 b=4$$

Rewrite 5r^{2}-11r-12 as \left(5r^{2}-15r\right)+\left(4r-12\right).

$$\left(5r^{2}-15r\right)+\left(4r-12\right)$$

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

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

Factor out common term r-3 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 11r from both sides.

$$5r^{2}-11r=12$$

Subtract 12 from both sides.

$$5r^{2}-11r-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -11 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 5\left(-12\right)}}{2\times 5}$$

Square -11.

$$r=\frac{-\left(-11\right)±\sqrt{121-4\times 5\left(-12\right)}}{2\times 5}$$

Multiply -4 times 5.

$$r=\frac{-\left(-11\right)±\sqrt{121-20\left(-12\right)}}{2\times 5}$$

Multiply -20 times -12.

$$r=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 5}$$

Add 121 to 240.

$$r=\frac{-\left(-11\right)±\sqrt{361}}{2\times 5}$$

Take the square root of 361.

$$r=\frac{-\left(-11\right)±19}{2\times 5}$$

The opposite of -11 is 11.

$$r=\frac{11±19}{2\times 5}$$

Multiply 2 times 5.

$$r=\frac{11±19}{10}$$

Now solve the equation r=\frac{11±19}{10} when ± is plus. Add 11 to 19.

$$r=\frac{30}{10}$$

Divide 30 by 10.

$$r=3$$

Now solve the equation r=\frac{11±19}{10} when ± is minus. Subtract 19 from 11.

$$r=-\frac{8}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 11r from both sides.

$$5r^{2}-11r=12$$

Divide both sides by 5.

$$\frac{5r^{2}-11r}{5}=\frac{12}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$r^{2}-\frac{11}{5}r=\frac{12}{5}$$

Divide -\frac{11}{5}, the coefficient of the x term, by 2 to get -\frac{11}{10}. Then add the square of -\frac{11}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-\frac{11}{5}r+\left(-\frac{11}{10}\right)^{2}=\frac{12}{5}+\left(-\frac{11}{10}\right)^{2}$$

Square -\frac{11}{10} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-\frac{11}{5}r+\frac{121}{100}=\frac{12}{5}+\frac{121}{100}$$

Add \frac{12}{5} to \frac{121}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}-\frac{11}{5}r+\frac{121}{100}=\frac{361}{100}$$

Factor r^{2}-\frac{11}{5}r+\frac{121}{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}.

$$\left(r-\frac{11}{10}\right)^{2}=\frac{361}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{11}{10}\right)^{2}}=\sqrt{\frac{361}{100}}$$

Simplify.

$$r-\frac{11}{10}=\frac{19}{10} r-\frac{11}{10}=-\frac{19}{10}$$

Add \frac{11}{10} to both sides of the equation.

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