3q^2-20q-100=0

Solution for 3q^2-20q-100=0 equation:

3q^2-20q-100=0

a = 3; b = -20; c = -100;
Δ = b2-4ac
Δ = -202-4·3·(-100)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-40}{2*3}=\frac{-20}{6}
=-3+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+40}{2*3}=\frac{60}{6}
=10
$