-3q2+80q-170=0

Solution for -3q2+80q-170=0 equation:

-3q^2+80q-170=0

a = -3; b = 80; c = -170;
Δ = b2-4ac
Δ = 802-4·(-3)·(-170)
Δ = 4360
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4360}=\sqrt{4*1090}=\sqrt{4}*\sqrt{1090}=2\sqrt{1090}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-2\sqrt{1090}}{2*-3}=\frac{-80-2\sqrt{1090}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+2\sqrt{1090}}{2*-3}=\frac{-80+2\sqrt{1090}}{-6}
$