\(x^2-4x-3=0\)
Theo Vi-ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=4\\x_1x_2=\dfrac{c}{a}=-3\end{matrix}\right.\)
Ta có :
\(B=3x_1^2+3x_2^2-5x_1x_2\)
\(=3\left(x_1^2+x_2^2\right)-5x_1x_2\)
\(=3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2\)
\(=3[4^2-2.\left(-3\right)]-5.\left(-3\right)\)
\(=81\)