40=6q+6q2=40

Solution for 40=6q+6q2=40 equation:

40=6q+6q^2=40

We move all terms to the left:

40-(6q+6q^2)=0

We get rid of parentheses

-6q^2-6q+40=0

a = -6; b = -6; c = +40;
Δ = b2-4ac
Δ = -62-4·(-6)·40
Δ = 996
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{996}=\sqrt{4*249}=\sqrt{4}*\sqrt{249}=2\sqrt{249}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{249}}{2*-6}=\frac{6-2\sqrt{249}}{-12}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{249}}{2*-6}=\frac{6+2\sqrt{249}}{-12}
$