0=y^2-12y-4

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

0=y^2-12y-4

We move all terms to the left:

0-(y^2-12y-4)=0

We add all the numbers together, and all the variables

-(y^2-12y-4)=0

We get rid of parentheses

-y^2+12y+4=0

We add all the numbers together, and all the variables

-1y^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}
$