2y+98=y^2-2y

Solution for 2y+98=y^2-2y equation:

2y+98=y^2-2y

We move all terms to the left:

2y+98-(y^2-2y)=0

We get rid of parentheses

-y^2+2y+2y+98=0

We add all the numbers together, and all the variables

-1y^2+4y+98=0

a = -1; b = 4; c = +98;
Δ = b2-4ac
Δ = 42-4·(-1)·98
Δ = 408
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:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{102}}{2*-1}=\frac{-4-2\sqrt{102}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{102}}{2*-1}=\frac{-4+2\sqrt{102}}{-2}
$