3q^2-4q-20=0

Solution for 3q^2-4q-20=0 equation:

3q^2-4q-20=0

a = 3; b = -4; c = -20;
Δ = b2-4ac
Δ = -42-4·3·(-20)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*3}=\frac{-12}{6}
=-2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*3}=\frac{20}{6}
=3+1/3
$