3q^2+10q-8000=0

Solution for 3q^2+10q-8000=0 equation:

3q^2+10q-8000=0

a = 3; b = 10; c = -8000;
Δ = b2-4ac
Δ = 102-4·3·(-8000)
Δ = 96100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{96100}=310$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-310}{2*3}=\frac{-320}{6}
=-53+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+310}{2*3}=\frac{300}{6}
=50
$