Answer
$p= -3.95, p= 2.28$
Work Step by Step
Use the quadratic formula. $$
\begin{aligned}
& (3 p-4)(p+3)=15 \\
& (3 p-4) p+3(3 p-4)=15 \\
& 3 p^2-4 p+9 p-12=15 \\
& 3 p^2+5 p-12=15 \\
& 3 p^2+5 p-12-15=0 \\
& 3 p^2+5 p-27=0.
\end{aligned}
$$ Set $$
\begin{aligned}
a & =3 \\
b& =5\\
d & =-27 \\
\end{aligned}
$$ $$
\begin{aligned}
x&= \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
x& =\frac{-(5) \pm \sqrt{(5)^2-4 \cdot 3 \cdot (-27)}}{2 \cdot 3 }\\
& =\frac{-5 \pm \sqrt{349}}{6}.
\end{aligned}
$$ This gives $$
\begin{aligned}
p & =\frac{-5-\sqrt{349}}{6}\\
& \approx-3.95\\
p & =\frac{-5+\sqrt{349}}{6} \\
& \approx 2.28
\end{aligned}
$$ Define the following function and plot it. We see that the zeros are where they should be.
$$
f(p) =3 p^2+5 p-27.
$$
