Answer
$x=\left\{
-\dfrac{1}{2},3
\right\}$
Work Step by Step
In the given equation,
\begin{align*}
2x^2-5x-3=0
,\end{align*} $a=
2
,$ $b=
-5
,$ and $c=
-3
.$
Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}x&=
\dfrac{-(-5)\pm\sqrt{(-5)^2-4(2)(-3)}}{2(2)}
\\\\&=
\dfrac{5\pm\sqrt{25+24}}{4}
\\\\&=
\dfrac{5\pm\sqrt{49}}{4}
\\\\&=
\dfrac{5\pm7}{4}
\end{align*}
\begin{array}{lcl}
\Rightarrow\dfrac{5-7}{4} &\text{OR}& \dfrac{5+7}{4}
\\\\
=\dfrac{-2}{4} && =\dfrac{12}{4}
\\\\
=-\dfrac{1}{2} && =3
.\end{array}
Hence, the solutions are $
x=\left\{
-\dfrac{1}{2},3
\right\}
.$
