In problems 7-22, solve the inequality.

$2x^2<5x+3$

The given inequality is$2x^2<5x+3$.

• Rewrite the given inequality.

$2x^2-5x-3<0$

• The value of $f\left(0\right)=-3$.
• So, the y-intercept is $(0,-3)$.
• Factor the equation.

role="math" localid="1646475257818" $\begin{array}{rcl}2x^2-5x-3& =& 0\\ 2x^2-6x+x-3& =& 0\\ 2x(x-3)+1(x-3)& =& 0\\ (2x+1)(x-3)& =& 0\end{array}$

• Equate the factors to 0.

role="math" localid="1646475273071" $\begin{array}{rcl}2x+1& =& 0\\ x& =& \frac{-1}{2}\\ x-3& =& 0\\ x& =& 3\end{array}$

• So, the x-intercepts are role="math" localid="1646475288022" $(\frac{-1}{2},0),(3,0)$.
• The vertex of the parabola exists at $\frac{-b}{2a}=\frac{-(-5)}{2\left(2\right)}=\frac{5}{4}$.
• Find the value of the function at $x=\frac{5}{4}$.

$\begin{array}{rcl}f\left(\frac{5}{4}\right)& =& 2{\left(\frac{5}{4}\right)}^2-5\left(\frac{5}{4}\right)-3\\ & =& 2\left(\frac{25}{16}\right)-\frac{25}{4}-3\\ & =& \frac{50-100-48}{16}\\ & =& \frac{-98}{16}\\ & =& \frac{-49}{8}\end{array}$

• So, the vertex is $(\frac{5}{4},\begin{array}{rcl}& & \frac{-49}{8}\end{array})$.

Plot the parabola on a graph using the vertex and intercepts calculated in the previous step.

• From the graph, it is observed that for x belonging to $(\frac{-1}{2},3)$, $y<0$.
• So, the solution set of the given inequality is $(\frac{-1}{2},3)$.