Solve -9x+6x^2

$$-9x+6 { x }^{ 2 }$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(-3x+2x^{2}\right)$$

Consider -3x+2x^{2}. Factor out x.

$$x\left(-3+2x\right)$$

Rewrite the complete factored expression.

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

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.

$$6x^{2}-9x=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}}}{2\times 6}$$

Take the square root of \left(-9\right)^{2}.

$$x=\frac{-\left(-9\right)±9}{2\times 6}$$

The opposite of -9 is 9.

$$x=\frac{9±9}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{9±9}{12}$$

Now solve the equation x=\frac{9±9}{12} when ± is plus. Add 9 to 9.

$$x=\frac{18}{12}$$

Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{9±9}{12} when ± is minus. Subtract 9 from 9.

$$x=\frac{0}{12}$$

Divide 0 by 12.

$$x=0$$

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

$$6x^{2}-9x=6\left(x-\frac{3}{2}\right)x$$

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

$$6x^{2}-9x=6\times \frac{2x-3}{2}x$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6x^{2}-9x=3\left(2x-3\right)x$$

Evaluate

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