3q^2-10q+6=0

Solution for 3q^2-10q+6=0 equation:

3q^2-10q+6=0

a = 3; b = -10; c = +6;
Δ = b2-4ac
Δ = -102-4·3·6
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{7}}{2*3}=\frac{10-2\sqrt{7}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{7}}{2*3}=\frac{10+2\sqrt{7}}{6}
$