Solve 12r-18r^2

$$12 r - 18 r ^ { 2 }$$

Factor

$$6r\left(2-3r\right)$$

Solution Steps

Factor out 6.

$$6\left(2r-3r^{2}\right)$$

Consider 2r-3r^{2}. Factor out r.

$$r\left(2-3r\right)$$

Rewrite the complete factored expression.

$$6r\left(-3r+2\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-18r^{2}+12r=0$$

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.

$$r=\frac{-12±\sqrt{12^{2}}}{2\left(-18\right)}$$

Take the square root of 12^{2}.

$$r=\frac{-12±12}{2\left(-18\right)}$$

Multiply 2 times -18.

$$r=\frac{-12±12}{-36}$$

Now solve the equation r=\frac{-12±12}{-36} when ± is plus. Add -12 to 12.

$$r=\frac{0}{-36}$$

Divide 0 by -36.

$$r=0$$

Now solve the equation r=\frac{-12±12}{-36} when ± is minus. Subtract 12 from -12.

$$r=-\frac{24}{-36}$$

Reduce the fraction \frac{-24}{-36} to lowest terms by extracting and canceling out 12.

$$r=\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and \frac{2}{3} for x_{2}.

$$-18r^{2}+12r=-18r\left(r-\frac{2}{3}\right)$$

Subtract \frac{2}{3} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-18r^{2}+12r=-18r\times \frac{-3r+2}{-3}$$

Cancel out 3, the greatest common factor in -18 and -3.

$$-18r^{2}+12r=6r\left(-3r+2\right)$$

Evaluate

$$6r\left(2-3r\right)$$