Question

Two consecutive integers whose product is 56

Answer 1

≡ We know that:
⇔ Consider the first integer is $a$
⇔ So, the next integer is $(a+1)$

≡ Solution:
⇒ $(a).(a+1)=56$
⇒ $a^{2}+a=56$
⇒ $a^{2}+a-56=0$
⇒ $(a+8)(a-7)=0$
⇒ $a_{1}=-8$║$a_{2}=7$

⇔  For $a=-8$
⇒ the next integer is $(a+1)=-8+1=-7$

⇔ For $a=7$
⇒ the next integer is $(a+1)=(7+1)=8$

∴ So, there are 2 options [$\boxed{-8}$ and $\boxed{-7}$] or [$\boxed{7}$ and $\boxed{8}$]

Answer 2

Let the two consecutive numbers be
x and x + 1

ATQ
X (x+1) = 55
x^2 + x = 56
x^2 + x -56 = 0
x^2 + 8x - 7x -56 = 0
x(x+8) - 7(x+8)=0
(x-7)(x+8) =0

x= 7 or x = -8