6q^2-40q+10=0

Solution for 6q^2-40q+10=0 equation:

6q^2-40q+10=0

a = 6; b = -40; c = +10;
Δ = b2-4ac
Δ = -402-4·6·10
Δ = 1360
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:

$\sqrt{\Delta}=\sqrt{1360}=\sqrt{16*85}=\sqrt{16}*\sqrt{85}=4\sqrt{85}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{85}}{2*6}=\frac{40-4\sqrt{85}}{12}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{85}}{2*6}=\frac{40+4\sqrt{85}}{12}
$