Answer
$P\left(\dfrac{3}{2}\right)=0$
Work Step by Step
Substituting $x=\dfrac{3}{2}$ in $
P(x)=2x^2+5x-12
$, results to
\begin{align*}
P\left(\dfrac{3}{2}\right)&=
2\left(\dfrac{3}{2}\right)^2+5\left(\dfrac{3}{2}\right)-12
\\\\&=
2\left(\dfrac{9}{4}\right)+5\left(\dfrac{3}{2}\right)-12
\\\\&=
\dfrac{9}{2}+\dfrac{15}{2}-12
\\\\&=
\dfrac{24}{2}-12
\\\\&=
12-12
\\&=
0
.\end{align*}
Hence, $P\left(\dfrac{3}{2}\right)=0$.
