0.3q^2+2q-50=0

Solution for 0.3q^2+2q-50=0 equation:

0.3q^2+2q-50=0

a = 0.3; b = 2; c = -50;
Δ = b2-4ac
Δ = 22-4·0.3·(-50)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*0.3}=\frac{-10}{0.6}
=-16+0.4/0.6
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*0.3}=\frac{6}{0.6}
=10
$