Explanation:
given equation \(18+15 x-3 x^{2}=0\)
we divide through by \(-3\)
\begin{array}{l}
\frac{18}{-3}+\frac{15 x}{-3}-\frac{3 x^{2}}{-3}=\frac{0}{3} \\
-6-5 x+x^{2}=0 \\
x^{2}-5 x-6=0 \\
a=1 . b=-5, c=-6
\end{array}
If \(\alpha\) and \(\beta\) are the roots of the equation then sum of roots \(\alpha+\beta=\frac{-b}{a}=\frac{-(-5)}{1}=5\)
\begin{array}{l}
\text { product of root }=\frac{c}{a}=\frac{-6}{1}=-6 \\
\begin{aligned}
\alpha \beta-\alpha-\beta &=\alpha \beta-(\alpha+\beta) \\
&=-6-(5)=-6-5=-11
\end{aligned}
\end{array}