3q2-6q-50=0

Solution for 3q2-6q-50=0 equation:

3q^2-6q-50=0

a = 3; b = -6; c = -50;
Δ = b2-4ac
Δ = -62-4·3·(-50)
Δ = 636
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:

$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{159}}{2*3}=\frac{6-2\sqrt{159}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{159}}{2*3}=\frac{6+2\sqrt{159}}{6}
$