3+10q-3q2=0

Solution for 3+10q-3q2=0 equation:

3+10q-3q^2=0

a = -3; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·(-3)·3
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{34}}{2*-3}=\frac{-10-2\sqrt{34}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{34}}{2*-3}=\frac{-10+2\sqrt{34}}{-6}
$