2y+98=y^2+34

Solution for 2y+98=y^2+34 equation:

2y+98=y^2+34

We move all terms to the left:

2y+98-(y^2+34)=0

We get rid of parentheses

-y^2+2y-34+98=0

We add all the numbers together, and all the variables

-1y^2+2y+64=0

a = -1; b = 2; c = +64;
Δ = b2-4ac
Δ = 22-4·(-1)·64
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{65}}{2*-1}=\frac{-2-2\sqrt{65}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{65}}{2*-1}=\frac{-2+2\sqrt{65}}{-2}
$