Question

the set of all natural number 'x' such that x2+5x+6=0​

Answer 1

$\large{\sf{\red{SolutiOn-}}}$

x^2+5x+6 can be factored into (x+2)(x+3). Since x is an integer, x+2 and x+3 must be as well. This means that either x+2 or x+3 must equal 2, or else it is impossible for the initial expression to be equal to a prime number. Thus, our two possibilities for x are -1 and -2, with a corresponding x^2+5x+6 value of 2 or 0. Since prime numbers are defined to be positive, the x=-2 case can be excluded, yielding an answer of x = -1.

I realised that I forgot about the case where x+3 or x+2 is -1, in which case, following a similar strand of reasoning, x=-4 presents itself as another possible solution.

Answer 2

Step-by-step explanation:

2x + 5x +6 = 0

7x = -6

x =-6 /7

if you want to say

x² + 5x +6=0

than

x² +3x +2x +6

x (x + 3 ) 2 ( x +3)

(x+2) (x+ 3)

than x have 2 value

x= -2

and

x = -3