Answer
$x=\left\{ \dfrac{-3-\sqrt{17}}{4},\dfrac{-3+\sqrt{17}}{4} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
2x^2+3x-1=0
,$ use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the form $ax^2+bx+c=0,$ the quadratic equation above has $a=
2
, b=
3
, c=
-1
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-3\pm\sqrt{3^2-4(2)(-1)}}{2(2)}
\\\\
x=\dfrac{-3\pm\sqrt{9+8}}{4}
\\\\
x=\dfrac{-3\pm\sqrt{17}}{4}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-3-\sqrt{17}}{4}
\\\\\text{OR}\\\\
x=\dfrac{-3+\sqrt{17}}{4}
.\end{array}
Hence, $
x=\left\{ \dfrac{-3-\sqrt{17}}{4},\dfrac{-3+\sqrt{17}}{4} \right\}
.$
