y^2+20y=3500

Solution for y^2+20y=3500 equation:

y^2+20y=3500

We move all terms to the left:

y^2+20y-(3500)=0

a = 1; b = 20; c = -3500;
Δ = b2-4ac
Δ = 202-4·1·(-3500)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-120}{2*1}=\frac{-140}{2}
=-70
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+120}{2*1}=\frac{100}{2}
=50
$