0=2w^2+2w-840

Solution for 0=2w^2+2w-840 equation:

0=2w^2+2w-840

We move all terms to the left:

0-(2w^2+2w-840)=0

We add all the numbers together, and all the variables

-(2w^2+2w-840)=0

We get rid of parentheses

-2w^2-2w+840=0

a = -2; b = -2; c = +840;
Δ = b2-4ac
Δ = -22-4·(-2)·840
Δ = 6724
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6724}=82$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-82}{2*-2}=\frac{-80}{-4}
=+20
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+82}{2*-2}=\frac{84}{-4}
=-21
$