Solve -4x^2+10x

$$- 4 x ^ { 2 } + 10 x$$

Factor

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

Solution Steps

Factor out 2.

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

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

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

Rewrite the complete factored expression.

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

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.

$$-4x^{2}+10x=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}}}{2\left(-4\right)}$$

Take the square root of 10^{2}.

$$x=\frac{-10±10}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-10±10}{-8}$$

Now solve the equation x=\frac{-10±10}{-8} when ± is plus. Add -10 to 10.

$$x=\frac{0}{-8}$$

Divide 0 by -8.

$$x=0$$

Now solve the equation x=\frac{-10±10}{-8} when ± is minus. Subtract 10 from -10.

$$x=-\frac{20}{-8}$$

Reduce the fraction \frac{-20}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

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

$$-4x^{2}+10x=-4x\left(x-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+10x=-4x\times \frac{-2x+5}{-2}$$

Cancel out 2, the greatest common factor in -4 and -2.

$$-4x^{2}+10x=2x\left(-2x+5\right)$$

Evaluate

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