y^2+12y+1=16

Solution for y^2+12y+1=16 equation:

y^2+12y+1=16

We move all terms to the left:

y^2+12y+1-(16)=0

We add all the numbers together, and all the variables

y^2+12y-15=0

a = 1; b = 12; c = -15;
Δ = b2-4ac
Δ = 122-4·1·(-15)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{51}}{2*1}=\frac{-12-2\sqrt{51}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{51}}{2*1}=\frac{-12+2\sqrt{51}}{2}
$