Solve 6x+9x^2=

$$6 x + 9 x ^ { 2 } =$$

Factor

$$3x\left(3x+2\right)$$

Solution Steps

Factor out 3.

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

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

$$x\left(2+3x\right)$$

Rewrite the complete factored expression.

$$3x\left(3x+2\right)$$

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.

$$9x^{2}+6x=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}}}{2\times 9}$$

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-6±6}{18}$$

Now solve the equation x=\frac{-6±6}{18} when ± is plus. Add -6 to 6.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

Now solve the equation x=\frac{-6±6}{18} when ± is minus. Subtract 6 from -6.

$$x=-\frac{12}{18}$$

Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{2}{3}$$

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

$$9x^{2}+6x=9x\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$9x^{2}+6x=9x\left(x+\frac{2}{3}\right)$$

Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}+6x=9x\times \frac{3x+2}{3}$$

Cancel out 3, the greatest common factor in 9 and 3.

$$9x^{2}+6x=3x\left(3x+2\right)$$

Evaluate

$$3x\left(3x+2\right)$$