-3y+38=y^2

Solution for -3y+38=y^2 equation:

-3y+38=y^2

We move all terms to the left:

-3y+38-(y^2)=0

determiningTheFunctionDomain
-y^2-3y+38=0

We add all the numbers together, and all the variables

-1y^2-3y+38=0

a = -1; b = -3; c = +38;
Δ = b2-4ac
Δ = -32-4·(-1)·38
Δ = 161
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{161}}{2*-1}=\frac{3-\sqrt{161}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{161}}{2*-1}=\frac{3+\sqrt{161}}{-2}
$