Answer
$\left\{ \dfrac{-7-\sqrt{33}}{2},\dfrac{-7+\sqrt{33}}{2} \right\}$
Work Step by Step
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the given quadratic equation, $
x^2+7x+4=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(7)\pm\sqrt{(7)^2-4(1)(4)}}{2(1)}
\\\\=
\dfrac{-7\pm\sqrt{49-16}}{2}
\\\\=
\dfrac{-7\pm\sqrt{33}}{2}
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{-7-\sqrt{33}}{2},\dfrac{-7+\sqrt{33}}{2} \right\}
.$
