0=y^2-8y+12

Solution for 0=y^2-8y+12 equation:

0=y^2-8y+12

We move all terms to the left:

0-(y^2-8y+12)=0

We add all the numbers together, and all the variables

-(y^2-8y+12)=0

We get rid of parentheses

-y^2+8y-12=0

We add all the numbers together, and all the variables

-1y^2+8y-12=0

a = -1; b = 8; c = -12;
Δ = b2-4ac
Δ = 82-4·(-1)·(-12)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*-1}=\frac{-12}{-2}
=+6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*-1}=\frac{-4}{-2}
=+2
$