SOLUTION: how do i find how many real numbers the equation -4j^2+3j-28=0 has?

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Question 174885: how do i find how many real numbers the equation -4j^2+3j-28=0 has?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
To find the number of real solutions, we can use the discriminant formula.


From -4j%5E2%2B3j-28 we can see that a=-4, b=3, and c=-28


D=b%5E2-4ac Start with the discriminant formula.


D=%283%29%5E2-4%28-4%29%28-28%29 Plug in a=-4, b=3, and c=-28


D=9-4%28-4%29%28-28%29 Square 3 to get 9


D=9-448 Multiply 4%28-4%29%28-28%29 to get %28-16%29%28-28%29=448


D=-439 Subtract 448 from 9 to get -439


Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.