For what values of j does the equation (2x + 7)(x + 43) = -5 + jx have exactly one real solution? Express your answer as a list of numbers, separated by commas.
+36916 (2x + 7)(x + 43) = -5 + jx expands and simplifies to
2x^2 + 86x + 7x + 301 - jx + 5 = 0
2x^2 93x-jx + 306 = 0
If the disciminant = 0 then there is only one solution
(b^2) - 4ac = 0
(93-j)^2 - 4( 2)(306) = 0
8649 - 186j + j^2 - 2448 = 0
j^2 -186j + 6201 = 0 Use quadratic Formula to find j = 93+- 12 sqrt(17)
