$Δ=(-15)²-4.4.(-6)=321$
$→$ Pt có 2 nghiệm phân biệt
$x_1=\dfrac{15+\sqrtΔ}{2.4}=\dfrac{15+\sqrt{321}}{8}$
$x_2=\dfrac{15-\sqrtΔ}{2.4}=\dfrac{15-\sqrt{321}}{8}$
$→\begin{cases}x_1+x_2=\dfrac{15+\sqrt{321}+15-\sqrt{321}}{8}=\dfrac{15}{4}\\x_1.x_2=\dfrac{(15+\sqrt{321})(15-\sqrt{321})}{64}=-\dfrac{3}{2}\end{cases}$