0=0.3q^2+q-40

Solution for 0=0.3q^2+q-40 equation:

0=0.3q^2+q-40

We move all terms to the left:

0-(0.3q^2+q-40)=0

We add all the numbers together, and all the variables

-(0.3q^2+q-40)=0

We get rid of parentheses

-0.3q^2-q+40=0

We add all the numbers together, and all the variables

-0.3q^2-1q+40=0

a = -0.3; b = -1; c = +40;
Δ = b2-4ac
Δ = -12-4·(-0.3)·40
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*-0.3}=\frac{-6}{-0.6}
=+10
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*-0.3}=\frac{8}{-0.6}
=-13+0.2/0.6
$