y2+8y+4y=-6

Solution for y2+8y+4y=-6 equation:

y2+8y+4y=-6

We move all terms to the left:

y2+8y+4y-(-6)=0

We add all the numbers together, and all the variables

y^2+12y+6=0

a = 1; b = 12; c = +6;
Δ = b2-4ac
Δ = 122-4·1·6
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{30}}{2*1}=\frac{-12-2\sqrt{30}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{30}}{2*1}=\frac{-12+2\sqrt{30}}{2}
$