2^2+11y=y^2+3y-28

Solution for 2^2+11y=y^2+3y-28 equation:

2^2+11y=y^2+3y-28

We move all terms to the left:

2^2+11y-(y^2+3y-28)=0

We add all the numbers together, and all the variables

11y-(y^2+3y-28)+4=0

We get rid of parentheses

-y^2+11y-3y+28+4=0

We add all the numbers together, and all the variables

-1y^2+8y+32=0

a = -1; b = 8; c = +32;
Δ = b2-4ac
Δ = 82-4·(-1)·32
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{3}}{2*-1}=\frac{-8-8\sqrt{3}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{3}}{2*-1}=\frac{-8+8\sqrt{3}}{-2}
$