Solve 2x^2-5x

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

Factor

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

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.

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

Take the square root of \left(-5\right)^{2}.

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

The opposite of -5 is 5.

$$x=\frac{5±5}{2\times 2}$$

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.

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

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

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

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

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

Divide 0 by 4.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and 0 for x_{2}.

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

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

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

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

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

Evaluate

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