Answer
The solutions are $x = -\frac{1}{3}, 2$
Work Step by Step
Rewrite the equation so that all terms are on the left side of the equation:
$$3x^2 - 5x - 2 = 0$$
To factor a quadratic trinomial in the form $ax^2 + bx + c0$, we look for the factors of the product of $a$ and $c$ that when added together equals $b$.
For the trinomial $3x^2 - 5x - 2$, $(a)(c)$ is $(3)(-2)$, or $-6$, but 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.
We came up with the following possibilities:
$-6=(-6)(1)$
$-6+1 = -5$
$-6=(-3)(2)$
$-3+2 = -1$
The first pair, $-6$ and $1$, is the one we are looking for.
Use these factors to split the middle term of the trinomial:
$$3x^2 - 6x + x - 2 = 0$$
Factor by grouping. Group the first two terms together and the last two terms together:
$$(3x^2 - 6x) + (x - 2) = 0$$
$3x$ is a common factor for the first group, so factor it out to obtain:
$$3x(x - 2) + (x - 2)$$
$x - 2$ is a common factor of both groups, factor it out to obtain:
$$(3x + 1)(x - 2)$$
Thus, the given equation is equivalent to:
$$(3x+1)(x-2)=0$$
Use the Zero-Product Property by equating each factor to zero, then solve each one.
For the first factor:
$3x + 1 = 0$
$3x = -1$
Divide each side by $3$ to solve for $x$:
$x = -\frac{1}{3}$
For the second factor:
$x - 2 = 0$
$x = 2$
The solutions are $x = -\frac{1}{3}, 2$.
To check if our solutions are correct, substitute each solution into the equation.
When $x = -\frac{1}{3}$:
$3x^2-2=5x\\
3\left(-\frac{1}{3}\right)^2 -2= 5\left(-\frac{1}{3}\right)\\
3\left(\frac{1}{9}\right)-2=-\frac{5}{3}\\
\frac{1}{3}-2=-\frac{5}{3}\\
-\frac{1}{3}-\frac{6}{3}=-\frac{5}{3}\\
-\frac{5}{3}=-\frac{5}{3}$
Thus, the solutions are $-\frac{1}{3}$ and $2$.
