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

Solution for 3(q)^2-80q+170=0 equation:

3(q)^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}
$