Wiemy, że:

$\frac{x_1+x_2}{2}=3\wedge q=36$  

zatem:  

$p=3\wedge q=36$  

zatem:

$\{\begin{array}{l}p=3\\ q=36\end{array}$  

$\{\begin{array}{l}-\frac{b}{2a}=3\\ -\frac{\Delta }{4}a=36\end{array}$  

$\{\begin{array}{l}-b=6a\\ -\Delta =144a\end{array}$  

$\{\begin{array}{l}-b=6a\\ -\left(b^2-4ac\right)=144a\end{array}$

$\{\begin{array}{l}-b=6a\\ -\left(b^2-4a\cdot 0\right)=144a\end{array}$

$\{\begin{array}{l}b=-6a\\ -{\left(-6a\right)}^2=144a\end{array}$

$\{\begin{array}{l}b=-6a\\ -36a^2=144a\end{array}$

$\{\begin{array}{l}b=-6a\\ -36a^2-144a=0\end{array}$

$\{\begin{array}{l}b=-6a\\ -36a^2-144a=0\mid :\left(-36\right)\end{array}$

$\{\begin{array}{l}b=-6a\\ a^2+4a=0\end{array}$

Rozwiążmy równanie:

$a^2+4a=0$  

$a\left(a+4\right)=0$  

$a=0\vee a+4=0$  

$a=0\vee a=-4$  

Funkcja f to funkcja kwadratowa, zatem:

$a=-4$  

czyli:

$b=-6\cdot \left(-4\right)=24$