[tex]\text{The quadratic formula:}\\\\ax^2+bx+c=0\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\--------------------------------\\\text{We have}\ -5x^2-12x+8=0\\\\a=-5,\ b=-12,\ c=8\\\\\text{Substitute:}\\\\b^2-4ac=(-12)^2-4(-5)(8)=144+160=304\\\\\sqrt{b^2-4ac}=\sqrt{304}=\sqrt{16\cdot9}=\sqrt{16}\cdot\sqrt{19}=4\sqrt{19}\\\\x_1=\dfrac{-(-12)-4\sqrt{19}}{2(-5)}=\dfrac{12-4\sqrt{19}}{-10}=\boxed{-\dfrac{6-2\sqrt{19}}{5}}\\\\x_2=\dfrac{-(-12)+4\sqrt{19}}{2(-5)}=\dfrac{12+4\sqrt{19}}{-10}=\boxed{-\dfrac{6+2\sqrt{19}}{5}}[/tex]
