Answer
$\left\{\dfrac{3\pm\sqrt{17}}{4}\right\}$
Work Step by Step
Using $ax^2+bx+c=0,$ the given equation, $
2x^2-3x-1=0
,$ has $a=
2
$, $b=
-3
$ and $c=
-1
$. Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}
x&=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(2)(-1)}}{2(2)}
\\\\&=
\dfrac{3\pm\sqrt{9+8}}{4}
\\\\&=
\dfrac{3\pm\sqrt{17}}{4}
.\end{align*}
Hence, the solution set of $
2x^2-3x-1=0
$ is $
\left\{\dfrac{3\pm\sqrt{17}}{4}\right\}
$.
