Solve h(x)=-2x^2+5x

$$h ( x ) = - 2 x ^ { 2 } + 5 x$$

Factor

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

Solution Steps

Factor out x.

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

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.

$$-2x^{2}+5x=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}}}{2\left(-2\right)}$$

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-5±5}{-4}$$

Now solve the equation x=\frac{-5±5}{-4} when ± is plus. Add -5 to 5.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-5±5}{-4} when ± is minus. Subtract 5 from -5.

$$x=-\frac{10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

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

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}.

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

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

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

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

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

Evaluate

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