Can a quadratic polynomial equation with real coefficients have one real solution and one complex solution? If so,

I think the rule to follow is to multiply the first number by the second and then add the first number. For example:
in the case of the first one, 1 x 4 = 4, then 4 + 1 = 5
for the second one, 2 x 5 = 10, and then 10 + 2 = 12...
So for 8+11:
simply multiply 8 x 11 = 88, and then add 88 + 8 = 96

The aggregate expense for the markers is 12.