Perpendicular equation to 4x+3y=12
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Let the two numbers be x and y, then,
xy = -12 . . . (1)
x + y = -10 . . . (2)
From (2), x = -10 - y . . . (3)
Putting (3) into (1), gives
(-10 - y)y = -12
-10y - y^2 = -12
y^2 + 10y - 12 = 0
Therefore, the two numbers are
and 
xy = -12 . . . (1)
x + y = -10 . . . (2)
From (2), x = -10 - y . . . (3)
Putting (3) into (1), gives
(-10 - y)y = -12
-10y - y^2 = -12
y^2 + 10y - 12 = 0
Therefore, the two numbers are