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.
(2x + 7) ( x - 5) = -43 + jx expand
2x^2 -8x - 35 = -43 + jx rearrange as
2x^2 - ( 8 + j)x + 8 = 0
This will have one real root if the discriminant = 0 ....so....
(8 + j)^2 - 4(2) ( 8) = 0
j^2 + 16j + 9 - 64 = 0
j^2 + 16j - 55 = 0
j = -8 + sqrt(119), -8 - sqrt(119)
+335 
