When a line in a slope graph has no rise/run like this one, there is no slope.
The slope is "Undefined".
The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.
In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.
So, the only feasible solutions are
For the record, in this case, none of the feasible solutions are actually a root of the polynomial.
Answer: -2
Step-by-step explanation: The line has a slope of -2/1, which is just -2.
Then, now that we have solved for T, we can evaluate and solve for t=20 minutes.