Answer
The solutions are $x = -\dfrac{1}{2}$ and $x= 3$.
Work Step by Step
To factor a quadratic polynomial in the form $ax^2 + bx + c$, we look at factors of the $(a)(c)$ term such that, when added together, equal the $b$ term.
For the expression $2x^2 - 5x - 3$, we look for the factors that, when multiplied together, will equal $(a)(c)$, which is $(2)(-3)$ or $-6$, but when adding the factors together will equal $b$ or $-5$:
$(a)(c)$ = $(-6)(1)$
$b = -5$
$(a)(c)$ = $(-3)(2)$
$b = 8$
The first pair works. We will use that pair to split the middle term:
$2x^2 - 6x + x - 3=0$
Now, we can factor by grouping. We group the first two terms together and the second two terms together:
$(2x^2 - 6x) + (x - 3)=0$
We see that $2x$ is a common factor for the first group, so let us factor out the $2x$ from that group:
$2x(x - 3) + (x - 3)=0$
We see that $x + 3$ is common to both groups, so we put that binomial in parentheses. The other binomial will be $3x - 1$, which is composed of the other numbers that are left (the $1$ is implied in front of the second group). We now have the two factors:
$(2x + 1)(x - 3) = 0$
Use the Zero-Product Property by equating each factor to zero, then solve each equation.
First factor:
$2x + 1 = 0$
$2x = -1$
Divide each side by $2$ to solve for $x$:
$x = -\frac{1}{2}$
Let us set the other factor equal to zero:
$x - 3 = 0$
$x = 3$
The solutions are $x = -\frac{1}{2}, 3$.
To check to see if our solution is correct, we plug in the values we got for $x$ into the original equation to see if both sides equal one another.
Let us test $-\frac{1}{2}$ first:
$2\left(-\frac{1}{2}\right)^2 - 5\left(-\frac{1}{2}\right) - 3 = 0$
$2\left(\frac{1}{4}\right) + \frac{5}{2} - 3 = 0$
$\frac{2}{4} + \frac{5}{2} - 3 = 0$
$\frac{1}{2} + \frac{5}{2} - 3 = 0$
$\frac{6}{2} - 3 = 0$
$3 - 3 = 0$
$0 = 0$
This solution is correct.
Now, let us look at the solution $3$:
$2(3^2) - 5(3) - 3 = 0$
$18 - 15 - 3 = 0$
$3 - 3 = 0$
$0 = 0$
This solution is also correct.
