4y+y^2=60

Solution for 4y+y^2=60 equation:

4y+y^2=60

We move all terms to the left:

4y+y^2-(60)=0

a = 1; b = 4; c = -60;
Δ = b2-4ac
Δ = 42-4·1·(-60)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*1}=\frac{-20}{2}
=-10
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*1}=\frac{12}{2}
=6
$