There are no integer solutions, but there are complex (imaginary number) solutions, where y2 - 6y +10 = 0 for y and x = 6-y
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You have two equations:
x + y = 6
x * y = 10
x= 6-y and sub in
y*(6-y) = 10
y^2 - 6y + 10 = 0 (quadratic equation)
y = -b +/- sqrt(b^2-4ac)/2a
= 6/2 +/- sqrt(36-4*1*10) / 2
= 6/2 +/- imag * sqrt(4)/2
= 3 +/- imag * 1
number 1= 3 + i
number 2 = 3 - i
14
34
260
-144
175
-56
-56
-1
34
-10 and -3
7 & 3
-56