3y+89=y^2-9

Solution for 3y+89=y^2-9 equation:

3y+89=y^2-9

We move all terms to the left:

3y+89-(y^2-9)=0

We get rid of parentheses

-y^2+3y+9+89=0

We add all the numbers together, and all the variables

-1y^2+3y+98=0

a = -1; b = 3; c = +98;
Δ = b2-4ac
Δ = 32-4·(-1)·98
Δ = 401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{401}}{2*-1}=\frac{-3-\sqrt{401}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{401}}{2*-1}=\frac{-3+\sqrt{401}}{-2}
$