-3y+28=y2

Solution for -3y+28=y2 equation:

-3y+28=y2

We move all terms to the left:

-3y+28-(y2)=0

We add all the numbers together, and all the variables

-1y^2-3y+28=0

a = -1; b = -3; c = +28;
Δ = b2-4ac
Δ = -32-4·(-1)·28
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*-1}=\frac{-8}{-2}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*-1}=\frac{14}{-2}
=-7
$