math

\(x^2+tx-10 = (x+a)(x+b)\\ x^2 + tx - 10 = x^2 + (a+b)x+ab\\ \therefore ab = -10\\ \text{For }a,b\in\mathbb Z,(a,b)=(10,-1),(5,-2),(2,-5),(1,-10),(-1,10),(-2,5),(-5,2),(-10,1)\\ \therefore t = 9,3,-3,-9\\ \text{Product of all t} = 9^2\cdot 3^2 = 729\)

I'm assuming that this is

(2/3)t - 1 < t + 7 ≤ 2t + 15

We have two inequalities

(2/3)t - 1 < t + 7                         and            t + 7 ≤ 2t + 15

-8 < (1/3)t                                                    -8 ≤ t 

-24 < t

Taking the most restrictive interval  we have that

[-8, inf )