0=2q2-18q+40

Solution for 0=2q2-18q+40 equation:

0=2q^2-18q+40

We move all terms to the left:

0-(2q^2-18q+40)=0

We add all the numbers together, and all the variables

-(2q^2-18q+40)=0

We get rid of parentheses

-2q^2+18q-40=0

a = -2; b = 18; c = -40;
Δ = b2-4ac
Δ = 182-4·(-2)·(-40)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*-2}=\frac{-20}{-4}
=+5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*-2}=\frac{-16}{-4}
=+4
$