Given that y is directly proportional to the $(3x + 7)^2$, and that $y = 6$ when $x = -4$.

1. Translate the direct proportion statement into an equation.

Since $y$ is directly proportional to $(3x+7)^2$,
$y = k(3x+7)^2$
where $k$ is a constant.

2. Substitute the given values to find the constant of proportionality.

Given $y=6$ when $x=-4$,
$6 = k(3(-4)+7)^2$
$6 = k(-12+7)^2$
$6 = 25k$
$k = \frac{6}{25}$

1. Substitute $y=15.36$ into the expression from part (a).

From part (a),
$y=6(3x+7)^2.$
When $y=15.36$,
$15.36=6(3x+7)^2$
$(3x+7)^2=\frac{15.36}{6}=2.56.$

2. Take square roots and solve for both possible values of $x$.

$3x+7=\pm\sqrt{2.56}=\pm1.6.$
So
$3x+7=1.6 \quad \text{or} \quad 3x+7=-1.6.$
For $3x+7=1.6$,
$3x=-5.4$
$x=-1.8.$
For $3x+7=-1.6$,
$3x=-8.6$
$x=-\frac{8.6}{3}=-\frac{43}{15}.$