So,
Given:
[tex]f(x)=8x^2+18x+5[/tex]We want to find the value of x when f(x) = -4.
For this, we could replace:
[tex]\begin{gathered} 8x^2+18x+5=-4 \\ 8x^2+18x+9=0 \end{gathered}[/tex]Now, let's solve this quadratic equation:
We could apply the quadratic formula as follows;
Given a quadratic equation of the form:
[tex]ax^2+bx+c=0[/tex]The solutions of this equation can be found using the following formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]If we replace the values of our equation, these are:
a=8
b=18
c=9
Thus,
[tex]\begin{gathered} x=\frac{-18\pm\sqrt[]{18^2-4(8)(9)}}{2(8)} \\ \\ x=\frac{-18\pm\sqrt[]{36}}{16}\to\begin{cases}x_1=\frac{-18+6}{16}=\frac{-12}{16}=-\frac{3}{4} \\ x_2=\frac{-18-6}{16}=\frac{-24}{16}=-\frac{3}{2}\end{cases} \end{gathered}[/tex]Therefore, the values of x are -3/2 and -3/4
