Solve 20x^2+15x

$$20 x ^ { 2 } + 15 x$$

Factor

$$5x\left(4x+3\right)$$

Solution Steps

Factor out 5.

$$5\left(4x^{2}+3x\right)$$

Consider 4x^{2}+3x. Factor out x.

$$x\left(4x+3\right)$$

Rewrite the complete factored expression.

$$5x\left(4x+3\right)$$

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.

$$20x^{2}+15x=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}}}{2\times 20}$$

Take the square root of 15^{2}.

$$x=\frac{-15±15}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-15±15}{40}$$

Now solve the equation x=\frac{-15±15}{40} when ± is plus. Add -15 to 15.

$$x=\frac{0}{40}$$

Divide 0 by 40.

$$x=0$$

Now solve the equation x=\frac{-15±15}{40} when ± is minus. Subtract 15 from -15.

$$x=-\frac{30}{40}$$

Reduce the fraction \frac{-30}{40} to lowest terms by extracting and canceling out 10.

$$x=-\frac{3}{4}$$

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{3}{4} for x_{2}.

$$20x^{2}+15x=20x\left(x-\left(-\frac{3}{4}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$20x^{2}+15x=20x\left(x+\frac{3}{4}\right)$$

Add \frac{3}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+15x=20x\times \frac{4x+3}{4}$$

Cancel out 4, the greatest common factor in 20 and 4.

$$20x^{2}+15x=5x\left(4x+3\right)$$

Evaluate

$$5x\left(4x+3\right)$$