0=y^2-42y+108

Solution for 0=y^2-42y+108 equation:

0=y^2-42y+108

We move all terms to the left:

0-(y^2-42y+108)=0

We add all the numbers together, and all the variables

-(y^2-42y+108)=0

We get rid of parentheses

-y^2+42y-108=0

We add all the numbers together, and all the variables

-1y^2+42y-108=0

a = -1; b = 42; c = -108;
Δ = b2-4ac
Δ = 422-4·(-1)·(-108)
Δ = 1332
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:

$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{37}}{2*-1}=\frac{-42-6\sqrt{37}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{37}}{2*-1}=\frac{-42+6\sqrt{37}}{-2}
$