We are looking for how far from x = 5 (the 5 in the x -> 5 expression) we can be and produce the value ε = 2.4 from the limit of 5 (the five on the RHS of the equation.)
So, we will substitute x = 5 + δ and L = 5 + ε = 5 + 2.4 into the given equation and solve for the desired δ.
x2 + x - 25 = 5
(5 + δ)2 + (5 + δ) - 25 = (5 + ε)
(25 + 10δ + δ2) + 5 + δ - 25 = 5 + (2.4)
δ2 + 11δ + 5 = 7.4
δ2 + 11δ - 2.4 = 0
Using the Quad Formula, δ = 0.2140179
Another way to think of the solution is that we can easily show that the limit as x -> 5 of the equation does indeed = 5 by substituting 5 into the equation to show that 5 = 5. In other words, (5, 5) is on the graph.
By calculating that δ = 0.2140179, that means that using x = 5 + δ = 5 + 0.2140179 in the LHS of the equation, we will produce a value of 7.4 = 5 + ε = 5 + 2.4, which is 2.4 above the limit of interest for this problem.
