Answer
$\left\{\dfrac{-7\pm\sqrt{97}}{8}\right\}$
Work Step by Step
Using $ax^2+bx+c=0$, the given equation $
4x^2+7x-3=0
$ has
\begin{align*}
a=4,b=7,\text{ and }c=-3
.\end{align*}
Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
x&=\dfrac{-7\pm\sqrt{7^2-4(4)(-3)}}{2(4)}
\\\\&=
\dfrac{-7\pm\sqrt{49+48}}{8}
\\\\&=
\dfrac{-7\pm\sqrt{97}}{8}
.\end{align*}
Hence, the solution set of the equation $
4x^2+7x-3=0
$ is $\left\{\dfrac{-7\pm\sqrt{97}}{8}\right\}$.
