0=y^2-28y+28

Solution for 0=y^2-28y+28 equation:

0=y^2-28y+28

We move all terms to the left:

0-(y^2-28y+28)=0

We add all the numbers together, and all the variables

-(y^2-28y+28)=0

We get rid of parentheses

-y^2+28y-28=0

We add all the numbers together, and all the variables

-1y^2+28y-28=0

a = -1; b = 28; c = -28;
Δ = b2-4ac
Δ = 282-4·(-1)·(-28)
Δ = 672
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:

$\sqrt{\Delta}=\sqrt{672}=\sqrt{16*42}=\sqrt{16}*\sqrt{42}=4\sqrt{42}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{42}}{2*-1}=\frac{-28-4\sqrt{42}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{42}}{2*-1}=\frac{-28+4\sqrt{42}}{-2}
$