Answer
$\bf{Graph (d)}$
Work Step by Step
We have $2(x+2)^2+5(x+2)-3=0$
This implies $[2(x+2)-1][(x+2)+3]=0$
Now, $2(x+2)-1=0$
or, $x=\dfrac{-3}{2}$
and $(x+2)+3=0$
or, $x=-5$
To find the $x$-intercepts, we will have to take $f(x)=0$
Therefore,
Our desired $x$-intercepts are: {$-5,\dfrac{-3}{2}$}; which represents the $\bf{Graph (d)}$
