Y+y^2=3984

Solution for Y+y^2=3984 equation:

+Y^2=3984

We move all terms to the left:

+Y^2-(3984)=0

a = 1; b = 0; c = -3984;
Δ = b2-4ac
Δ = 02-4·1·(-3984)
Δ = 15936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{15936}=\sqrt{64*249}=\sqrt{64}*\sqrt{249}=8\sqrt{249}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{249}}{2*1}=\frac{0-8\sqrt{249}}{2}
=-\frac{8\sqrt{249}}{2}
=-4\sqrt{249}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{249}}{2*1}=\frac{0+8\sqrt{249}}{2}
=\frac{8\sqrt{249}}{2}
=4\sqrt{249}
$