There are no two real numbers that do, but two complex numbers that do are: (3 + i√45) and (3 - i√45).
If the numbers multiply to -54 and sum to 6, then two numbers that do are: (3 + √63) ≈ 10.937 and (3 - √63) ≈ -4.937
If the numbers multiply to -54 and sum to -6, then two numbers that do are: (-3 + √63) ≈ 4.937 and (-3 - √63) ≈ -10.937
Let the numbers by 'm' & 'n'
Hence
m + n = 6
mn = 54
Let n = 54/n
Substitute
m + 54/m = 6
Multiply through by 'm'
m^2 + 54 = 6m
Form a quadratic equation
m^2 - 6m + 54 = 0
This will not factor
So use Quadratic Equa'n
m = { - - 6 +/-sqrt[(-6)^2 - 4(1)(54)]} / 2(1)
m = { 6 +/- sqrt(36 -216]} / 2
m = { 6 +/- sqrt[-180} /2
Since we cannot find the square root of a negative number, then the answer is unresolved, unles wee move into IMAGINARY NNumbers.
Hence sqrt(-180) = sqrt(-1) X sqrt(180)
The square root of '-1' is defined as 'i'.
Sqrt(180) = +/-13.4164....
Hence
m = {6 +/- 13.4164i } / 2
M = 3 +/- 6.7082 i
or for clarity
m = 3 + 6.7082 i & 3 - 6.7082 i .
The numbers are -10 and -70
-8 and -8
-3 and 6
15 and -3
-4 and -25
26
Two (or four) digits added together cannot equal 42. Two-digit numbers multiplied together cannot equal 82.
The numbers are -10 and -70
Urny ony
-8 and -8
-3 and 6
15 and -3
-4 and -25
95
-141
-8 and -4
-13