Solve 18x^2+10x

$$18 x ^ { 2 } + 10 x$$

Factor

$$2x\left(9x+5\right)$$

Solution Steps

Factor out 2.

$$2\left(9x^{2}+5x\right)$$

Consider 9x^{2}+5x. Factor out x.

$$x\left(9x+5\right)$$

Rewrite the complete factored expression.

$$2x\left(9x+5\right)$$

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.

$$18x^{2}+10x=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}}}{2\times 18}$$

Take the square root of 10^{2}.

$$x=\frac{-10±10}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-10±10}{36}$$

Now solve the equation x=\frac{-10±10}{36} when ± is plus. Add -10 to 10.

$$x=\frac{0}{36}$$

Divide 0 by 36.

$$x=0$$

Now solve the equation x=\frac{-10±10}{36} when ± is minus. Subtract 10 from -10.

$$x=-\frac{20}{36}$$

Reduce the fraction \frac{-20}{36} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{9}$$

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}{9} for x_{2}.

$$18x^{2}+10x=18x\left(x-\left(-\frac{5}{9}\right)\right)$$

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

$$18x^{2}+10x=18x\left(x+\frac{5}{9}\right)$$

Add \frac{5}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}+10x=18x\times \frac{9x+5}{9}$$

Cancel out 9, the greatest common factor in 18 and 9.

$$18x^{2}+10x=2x\left(9x+5\right)$$

Evaluate

$$2x\left(9x+5\right)$$