Solve 10x-15x^2=

$$10 x - 15 x ^ { 2 } =$$

Factor

$$5x\left(2-3x\right)$$

Solution Steps

Factor out 5.

$$5\left(2x-3x^{2}\right)$$

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

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

Rewrite the complete factored expression.

$$5x\left(-3x+2\right)$$

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.

$$-15x^{2}+10x=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}}}{2\left(-15\right)}$$

Take the square root of 10^{2}.

$$x=\frac{-10±10}{2\left(-15\right)}$$

Multiply 2 times -15.

$$x=\frac{-10±10}{-30}$$

Now solve the equation x=\frac{-10±10}{-30} when ± is plus. Add -10 to 10.

$$x=\frac{0}{-30}$$

Divide 0 by -30.

$$x=0$$

Now solve the equation x=\frac{-10±10}{-30} when ± is minus. Subtract 10 from -10.

$$x=-\frac{20}{-30}$$

Reduce the fraction \frac{-20}{-30} to lowest terms by extracting and canceling out 10.

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

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}.

$$-15x^{2}+10x=-15x\left(x-\frac{2}{3}\right)$$

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

$$-15x^{2}+10x=-15x\times \frac{-3x+2}{-3}$$

Cancel out 3, the greatest common factor in -15 and -3.

$$-15x^{2}+10x=5x\left(-3x+2\right)$$

Evaluate

$$5x\left(2-3x\right)$$