Answer
$x=\left\{ -\dfrac{16}{5},1 \right\}$
Work Step by Step
Let $z=(x+3)$. Then the given expression, $
5(x+3)^2-19(x+3)=4
,$ is equivalent to
\begin{array}{l}\require{cancel}
5z^2-19z=4
\\\\
5z^2-19z-4=0
\\\\
(5z+1)(z-4)=0
\\\\
z=\left\{ -\dfrac{1}{5}, 4 \right\}
.\end{array}
Since $z=x+3$, then if $z=-\dfrac{1}{5}$,
\begin{array}{l}\require{cancel}
-\dfrac{1}{5}=x+3
\\\\
-\dfrac{1}{5}-3=x
\\\\
x=-\dfrac{1}{5}-\dfrac{15}{5}
\\\\
x=-\dfrac{16}{5}
.\end{array}
Since $z=x+3$, then if $z=4$,
\begin{array}{l}\require{cancel}
4=x+3
\\\\
4-3=x
\\\\
x=1
.\end{array}
Hence, $
x=\left\{ -\dfrac{16}{5},1 \right\}
.$
