-3q2-80q-160=0

Solution for -3q2-80q-160=0 equation:

-3q^2-80q-160=0

a = -3; b = -80; c = -160;
Δ = b2-4ac
Δ = -802-4·(-3)·(-160)
Δ = 4480
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:

$\sqrt{\Delta}=\sqrt{4480}=\sqrt{64*70}=\sqrt{64}*\sqrt{70}=8\sqrt{70}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-8\sqrt{70}}{2*-3}=\frac{80-8\sqrt{70}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+8\sqrt{70}}{2*-3}=\frac{80+8\sqrt{70}}{-6}
$