Question

what is the factorisation of 2y^{2} - 38y - 228

Answer 1

2Y² - 38Y -228
=2(Y²-19Y-114)
GENERALLY YOU NEED TO FIND TWO NUMBERS SO THAT
1.THEIR SUM IS -19 
2.PRODUCT IS -114

BUT IN THIS CASE, IT IS NOT POSSIBLE.

SO FIRST FIND THE SOLUTION BY 
Y = $\frac{-(-38) + ((-38)^2-4*2*(-228)) ^1/2}{2*2}$ AND $\frac{-(-38) - ((-38)^2-4*2*(-228)) ^1/2}{2*2}$
OR Y = $\frac{1}{2}$(19 + √817) AND $\frac{1}{2}$(19 - √817)

NOW ONCE YOU GET THE SOLUTIONS, YOU CAN DIRECTLY WRITE THE FACTORIZATION AS 

2(Y - $\frac{1}{2}$(19 + √817))(Y - $\frac{1}{2}$(19 - √817))

Answer 2

2 (y² - 19 y - 114 ) 

114 = 3 * 2 * 19   - factorization does not seem to be possible directly with these factors.

roots are  = [ - b +- sqrt(b² - 4a c) ] / 2 
           = [19 +- sqrt(361 + 456) ] / 2 = (19+ √817 )/ 2  and (19-√817)/2

So three factors are : 2 ,  [ y - (19 + √817) /2 ] and  [ y - (19 - √817)/2 ]