For what values of j does the equation (2x+7)(x-5)=-43+jx have exactly one solution

Expand and rearrange the terms:
2x²-10x+7x-35=-43+jx
2x²-(3+j)x+8=0
x²-[(3+j)/2]x+4=0
for the equation to have exactly one solution, it has to be a square. Recall that (a-b)²=a²-2ab+b². In this case, a=x, b=2, 2ab=2*x*2=4x
so (3+j)/2=4
3+j=8
j=5

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