Answer
$x = -\frac{3}{2}, 1$
Work Step by Step
Rewrite the equation so that all terms are on the left side of the equation:
$$2x^2 + x - 3 = 0$$
To factor a quadratic trinomial in the form $ax^2 + bx + c$, we look for the factors of the product of $a$ and $c$ that when added together equals $b$.
For the trinomial $2x^2 + x - 3$, $(a)(c)$ is $(2)(-3)$, or $-6$, so we have to look for the factors of $-6$ that when added together will equal to $1$. One of the factors needs to be positive and the other one negative, but the positive 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:
$-6=(6)(-1)$
$6+(-1) = 5$
$-6=(3)(-2)$
$3+(-2) = 1$
The second pair of factors, $3$ and $-2$, is the one we are looking for.
Use $3$ and $-2$ to split the middle term:
$2x^2 + 3x - 2x - 3$
Factor by grouping. Group the first and third terms together and the second and fourth terms together:
$(2x^2 - 2x) + (3x - 3)$
$2x$ is a common factor of the first group and $3$ is a common factor of the second group. Factor them out to obtain:
$2x(x - 1) + 3(x - 1)$
$x - 1$ is a common factor of the two groups, factor it out to obtain:
$(2x + 3)(x - 1)$
Thus, the given equation is equivalent to:
$$(2x+3)(x-1)=0$$
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:
$x - 1 = 0$
$x = 1$
The solutions are $x = -\frac{3}{2}, 1$.
To check if the solutions are correct, substitute each value of $x$ into the equation:
When $x = -\frac{3}{2}$:
$2x^2+x=3\\
2\left(-\frac{3}{2}\right)^2 + \left(-\frac{3}{2}\right)=3\\
2\left(\frac{9}{4}\right)-\frac{3}{2}=3\\
\frac{9}{2}-\frac{3}{2}=3\\
\frac{6}{2}=3\\
3=3$
When $x = 1$:
$2x^2+x=3\\
2(1)^2 + 1=3\\
2(1)+1=3\\
2+1=3\\
3=3$
Therefore the solutions are $-\frac{3}{2}$ and $1$.
