Answer
See below:
Work Step by Step
Consider the inequality,
$2{{x}^{2}}+x-15<0$
The boundary points of the given inequality will be calculated by equating $f\left( x \right)$ to 0 as $2{{x}^{2}}+x-15=0$
Thus,
$\begin{align}
& 2{{x}^{2}}+6x-5x-15=0 \\
& 2x\left( x+3 \right)-5\left( x+3 \right)=0 \\
& \left( 2x-5 \right)\left( x+3 \right)=0
\end{align}$
Hence,
$x=-3,x=\frac{5}{2}$
From the above number line, the boundary points divide the number line into three parts,
$\left( -\infty ,-3 \right)\text{,}\left( -3,\frac{5}{2} \right)\text{ and }\left( \frac{5}{2},\infty \right)$
Now, one test value within each interval is chosen and f is evaluated at that number.
For both intervals $\left( -\infty ,-3 \right)\text{ and }\left( \frac{5}{2},\infty \right)$ , the function is positive.
And, for interval $\left( -3,\frac{5}{2} \right)$ the function is negative.
Also, the points $-3\text{ and }\frac{5}{2}$ are not included in the solution because the function is not equal to 0.
Hence, the required interval is $\left( -3,\frac{5}{2} \right)$
