12000=y^2+y

Solution for 12000=y^2+y equation:

12000=y^2+y

We move all terms to the left:

12000-(y^2+y)=0

We get rid of parentheses

-y^2-y+12000=0

We add all the numbers together, and all the variables

-1y^2-1y+12000=0

a = -1; b = -1; c = +12000;
Δ = b2-4ac
Δ = -12-4·(-1)·12000
Δ = 48001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{48001}}{2*-1}=\frac{1-\sqrt{48001}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{48001}}{2*-1}=\frac{1+\sqrt{48001}}{-2}
$