-w^2+21w=104

Solution for -w^2+21w=104 equation:

-w^2+21w=104

We move all terms to the left:

-w^2+21w-(104)=0

We add all the numbers together, and all the variables

-1w^2+21w-104=0

a = -1; b = 21; c = -104;
Δ = b2-4ac
Δ = 212-4·(-1)·(-104)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-5}{2*-1}=\frac{-26}{-2}
=+13
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+5}{2*-1}=\frac{-16}{-2}
=+8
$