y^2+12y-4=0

Solution for y^2+12y-4=0 equation:

y^2+12y-4=0

a = 1; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·1·(-4)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{10}}{2*1}=\frac{-12-4\sqrt{10}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{10}}{2*1}=\frac{-12+4\sqrt{10}}{2}
$