Solve 15x-25x^2

$$15x-25 { x }^{ 2 }$$

Factor

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

Solution Steps

Factor out 5.

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

Consider 3x-5x^{2}. Factor out x.

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

Rewrite the complete factored expression.

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

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.

$$-25x^{2}+15x=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}}}{2\left(-25\right)}$$

Take the square root of 15^{2}.

$$x=\frac{-15±15}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-15±15}{-50}$$

Now solve the equation x=\frac{-15±15}{-50} when ± is plus. Add -15 to 15.

$$x=\frac{0}{-50}$$

Divide 0 by -50.

$$x=0$$

Now solve the equation x=\frac{-15±15}{-50} when ± is minus. Subtract 15 from -15.

$$x=-\frac{30}{-50}$$

Reduce the fraction \frac{-30}{-50} to lowest terms by extracting and canceling out 10.

$$x=\frac{3}{5}$$

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{3}{5} for x_{2}.

$$-25x^{2}+15x=-25x\left(x-\frac{3}{5}\right)$$

Subtract \frac{3}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-25x^{2}+15x=-25x\times \frac{-5x+3}{-5}$$

Cancel out 5, the greatest common factor in -25 and -5.

$$-25x^{2}+15x=5x\left(-5x+3\right)$$

Evaluate

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