0=-2q^2+24q-40

Solution for 0=-2q^2+24q-40 equation:

0=-2q^2+24q-40

We move all terms to the left:

0-(-2q^2+24q-40)=0

We add all the numbers together, and all the variables

-(-2q^2+24q-40)=0

We get rid of parentheses

2q^2-24q+40=0

a = 2; b = -24; c = +40;
Δ = b2-4ac
Δ = -242-4·2·40
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16}{2*2}=\frac{8}{4}
=2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16}{2*2}=\frac{40}{4}
=10
$