There are three consecutive integers such that the square of the first

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

Then according to question

$x^{2}+(x+1)(x+2)=154$

$x^{2}+x^{2}+3 x+2=154$

$2 x^{2}+3 x+2-154=0$

$2 x^{2}+3 x-152=0$

$2 x^{2}+3 x-152=0$

$2 x^{2}-16 x+19 x-152=0$

$2 x(x-8)+19(x-8)=0$

$(x-8)(2 x+19)=0$

$(x-8)=0$

$x=8$

Or

$(2 x+19)=0$

$x=\frac{-19}{2}$

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

Therefore,

$x+1=8+1$

$=9$

And

$x+2=8+2$

$=10$

Thus, three consecutive positive integer be $8,9,10$