Answer
The solutions are $x = \frac{3}{2}, -\frac{2}{3}$.
Work Step by Step
Rewrite the equation so that all terms are on the left side of the equation:
$$6x^2 - 5x - 6 = 0$$
To factor a quadratic trinomial in the form $ax^2 + bx + c$, we look for factors of the product of $a$ and $c$ that when added together equals $b$.
For the trinomial $6x^2 - 5x - 6$, $(a)(c)$ is $(6)(-6)$, or $-36$, so look for the factors of $-36$ that when added together will equal $b$ or $-5$. One of the factors needs to be positive and the other one negative, but the negative factor should be the one with the greater absolute value. This is because a negative number multiplied with a positive number will equal a negative number; however, when a negative number is added to a positive number, the result can be either negative or positive, depending on which number has the greater absolute value.
The possibilities are:
$-36=(-36)(1)$
$-36+1 = -35$
$-36=(-12)(3)$
$-12+3 = -9$
$-36=(-9)(4)$
$-9+4 = -5$
The third pair, $-9$ and $4$, is the one we are looking for.
Use these factors to split the middle term:
$$6x^2 - 9x + 4x - 6$$
Factor by grouping. Group the first two terms together and the last two terms together:
$$(6x^2 - 9x) + (4x - 6)$$
$3x$ is a common factor for the first group, and $2$ is a common factor of the second group.
Factor them out to obtain:
$$3x(2x - 3) + 2(2x - 3)$$
$2x - 3$ is a common of both groups, so factor it out to obtain:
$$(2x - 3)(3x + 2)$$
Hence, the equation above is equivalent to:
$$(2x-3)(3x+2)=0$$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation.
For the first factor:
$2x - 3 = 0$
$2x = 3$
Divide each side by $2$ to solve for $x$:
$x = \frac{3}{2}$
For the second factor:
$3x + 2 = 0$
$3x = -2$
Divide each side by $3$ to solve for $x$:
$x = -\frac{2}{3}$
The solutions are $x = \frac{3}{2}, -\frac{2}{3}$.
To check if the solutions are correct, substitute each solution back into the equation.
When $x = \frac{3}{2}$:
$6x^2=5x+6\\
6\left(\frac{3}{2}\right)^2= 5\left(\frac{3}{2}\right) +6\\
6\left(\frac{9}{4}\right)=\frac{15}{2}+\frac{12}{2}\\
\frac{54}{4}=\frac{27}{2}\\
\frac{27}{2}=\frac{27}{2}$
When $x = -\frac{2}{3}$:
$6\left(-\frac{2}{3})\right)^{2}= 5\left(-\frac{2}{3}\right) +6\\
6\left(\frac{4}{9}\right)=-\frac{10}{3}+\frac{18}{3}\\
\frac{24}{9}=\frac{8}{3}\\
\frac{8}{3}=\frac{8}{3}$
Therefore the solutions are $\frac{3}{2}$ and $-\frac{2}{3}$.
