Solve -4x^2+5x

$$-4 { x }^{ 2 } +5x$$

Factor

$$x\left(5-4x\right)$$

Solution Steps

Factor out x.

$$x\left(-4x+5\right)$$

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.

$$-4x^{2}+5x=0$$

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.

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

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-5±5}{-8}$$

Now solve the equation x=\frac{-5±5}{-8} when ± is plus. Add -5 to 5.

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

Divide 0 by -8.

$$x=0$$

Now solve the equation x=\frac{-5±5}{-8} when ± is minus. Subtract 5 from -5.

$$x=-\frac{10}{-8}$$

Reduce the fraction \frac{-10}{-8} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{4}$$

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}{4} for x_{2}.

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

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Evaluate

$$x\left(5-4x\right)$$