Answer :
We need to solve the inequality:
[tex]x^{2}+3x-28<0[/tex]We can start by sketching the graph of the parabola:
[tex]x^{2}+3x-28=0[/tex]Then, we need to find the points of the parabola below the x-axis.
The zeros of that parabola are given by:
[tex]\begin{gathered} x=\frac{-3\pm\sqrt[]{3^{2}-4(1)(-28)}}{2(1)} \\ \\ x=\frac{-3\pm\sqrt[]{9+112}}{2} \\ \\ x=\frac{-3\pm\sqrt[]{121}}{2} \\ \\ x=\frac{-3\pm11}{2} \\ \\ x_1=-7 \\ \\ x_2=4 \end{gathered}[/tex]And since the coefficient of x² is positive, the parabola opens upwards. So, its graph looks like this:
Thus, the points on the parabola below the x-axis are those for which:
[tex]-7Therefore, in interval notation, the solution to the inequality is:[tex](-7,4)[/tex]