[tex]15x^2-40x-15 =0\ \ /:5\\ \\3x^2-8x-3 =0 \\ \\a=3, \ b=-8 , \ c= -3 \\ \\\Delta =b^2-4ac =(-8)^2 -4\cdot 3\cdot (-3)=64+36=100 \\ \\x_{1}=\frac{-b-\Delta }{2a}=\frac{8-\sqrt{100}}{2 \cdot 3}=\frac{8-10}{6}=-\frac{2}{6}=-\frac{1}{3}[/tex]
[tex]x_{2}=\frac{-b+\Delta }{2a}=\frac{8+\sqrt{100}}{2 \cdot 3}=\frac{8+10}{6}= \frac{18}{6}=3\\ \\ 3x^2-8x-3 = 3(x+\frac{1}{3})(x-3)=(3x+1)(x-3) \\ \\ \\Answer : \ 15x^2-40x-15 = 3x^2-8x-3 =(3x+1)(x-3)[/tex]
