Solve 20r^2+r=12

$$20 r ^ { 2 } + r = 12$$

Solve for r

$$r=\frac{3}{4}=0.75$$

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

Steps Using Factoring By Grouping

Subtract 12 from both sides.

$$20r^{2}+r-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20r^{2}+ar+br-12. To find a and b, set up a system to be solved.

$$a+b=1 ab=20\left(-12\right)=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 1.

$$a=-15 b=16$$

Rewrite 20r^{2}+r-12 as \left(20r^{2}-15r\right)+\left(16r-12\right).

$$\left(20r^{2}-15r\right)+\left(16r-12\right)$$

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

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

Factor out common term 4r-3 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$20r^{2}+r=12$$

Subtract 12 from both sides of the equation.

$$20r^{2}+r-12=12-12$$

Subtracting 12 from itself leaves 0.

$$20r^{2}+r-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 1 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-1±\sqrt{1^{2}-4\times 20\left(-12\right)}}{2\times 20}$$

Square 1.

$$r=\frac{-1±\sqrt{1-4\times 20\left(-12\right)}}{2\times 20}$$

Multiply -4 times 20.

$$r=\frac{-1±\sqrt{1-80\left(-12\right)}}{2\times 20}$$

Multiply -80 times -12.

$$r=\frac{-1±\sqrt{1+960}}{2\times 20}$$

Add 1 to 960.

$$r=\frac{-1±\sqrt{961}}{2\times 20}$$

Take the square root of 961.

$$r=\frac{-1±31}{2\times 20}$$

Multiply 2 times 20.

$$r=\frac{-1±31}{40}$$

Now solve the equation r=\frac{-1±31}{40} when ± is plus. Add -1 to 31.

$$r=\frac{30}{40}$$

Reduce the fraction \frac{30}{40} to lowest terms by extracting and canceling out 10.

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

Now solve the equation r=\frac{-1±31}{40} when ± is minus. Subtract 31 from -1.

$$r=-\frac{32}{40}$$

Reduce the fraction \frac{-32}{40} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

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

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$20r^{2}+r=12$$

Divide both sides by 20.

$$\frac{20r^{2}+r}{20}=\frac{12}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$r^{2}+\frac{1}{20}r=\frac{12}{20}$$

Reduce the fraction \frac{12}{20} to lowest terms by extracting and canceling out 4.

$$r^{2}+\frac{1}{20}r=\frac{3}{5}$$

Divide \frac{1}{20}, the coefficient of the x term, by 2 to get \frac{1}{40}. Then add the square of \frac{1}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+\frac{1}{20}r+\left(\frac{1}{40}\right)^{2}=\frac{3}{5}+\left(\frac{1}{40}\right)^{2}$$

Square \frac{1}{40} by squaring both the numerator and the denominator of the fraction.

$$r^{2}+\frac{1}{20}r+\frac{1}{1600}=\frac{3}{5}+\frac{1}{1600}$$

Add \frac{3}{5} to \frac{1}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}+\frac{1}{20}r+\frac{1}{1600}=\frac{961}{1600}$$

Factor r^{2}+\frac{1}{20}r+\frac{1}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r+\frac{1}{40}\right)^{2}=\frac{961}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+\frac{1}{40}\right)^{2}}=\sqrt{\frac{961}{1600}}$$

Simplify.

$$r+\frac{1}{40}=\frac{31}{40} r+\frac{1}{40}=-\frac{31}{40}$$

Subtract \frac{1}{40} from both sides of the equation.

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