If the equation has two integer solutions, then it can be written as \((x+\alpha)(x+\beta)\), where \(\alpha\) and \(\beta\) are integers.
Expanding this and setting it equal to \(x^2+bx+414720=0\), we find that \(b=\alpha+\beta\) and \(414720=\alpha\beta\).
414720 has 55 different pairs of factors. Each of these pairs leads to two values of \(b\) (a positive and a negative value). Therefore there are 110 values of \(b\) that give the equation two integer solutions.