Three consecutive positive integers are such that the sum

Let three consecutive integer be $x,(x+1)$ and $(x+2)$

Then according to question

$x^{2}+(x+1)(x+2)=46$

$x^{2}+x^{2}+3 x+2=46$

$2 x^{2}+3 x+2-46=0$

$2 x^{2}+3 x-44=0$

$2 x^{2}+3 x-44=0$

$2 x^{2}-8 x+11 x-44=0$

$2 x(x-4)+11(x-4)=0$

$(x-4)(2 x+11)=0$

$(x-4)=0$

$x=4$

Or

$(2 x+11)=0$

$x=\frac{-11}{2}$

Since, x being a positive number, so x cannot be negative.

Therefore,

When $x=4$ then other positive integer

$x+1=4+1$

$=5$

And

$x+2=4+2$

$=6$

Thus, three consecutive positive integer be $4,5,6$