Question -
Answer -
Let first number = x
Then second number = x + 1
and third number = x + 2
According co the condition,
(x)2 + (x+ 1) (x + 2) = 46
x2 + x2 + 3x + 2 = 46
=> 2x2 + 3x + 2 – 46 = 0
=> 2x2 + 3x – 44 = 0
=> 2x2 + 11x – 8x – 44 = 0
=> x (2x + 11) – 4 (2x + 11) = 0
=> (2x + 11) (x – 4) = 0
Either 2x + 11 = 0, then x = −11/2 which is not possible being fraction
or x – 4 = 0, then x = 4
Numbers are 4, 5, 6