For what values of $j$ does the equation $(2x+7)(x-5) = -43 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.
We can start by expanding the equation (2x + 7)(x - 5) = -43 + jx ---> 2x^2 - (j+3)x + 8 = 0.
Now we know that when the discriminant (b^2 - 4ac) is equal to 0, we have one real solution.
So we plug in our values and we get j^2 + 6j - 55 = 0.
Solving this gives us j = -11, 5