SOLUTION: how do i use the discriminant to determine how many real number solutions the quadratic equation -4j^2+3j-28=0 has?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: how do i use the discriminant to determine how many real number solutions the quadratic equation -4j^2+3j-28=0 has?      Log On


   

Question 253836: how do i use the discriminant to determine how many real number solutions the quadratic equation -4j^2+3j-28=0 has?
Answer by JimboP1977(311) About Me  (Show Source):
You can put this solution on YOUR website!
b%5E2-4ac is the discriminant.
b%5E2-4ac+%3E+0%29 2 real solutions
b%5E2-4ac+=+0%29 1 real solution
b%5E2-4ac+%3C+0%29 no real solutions
So in the case of -4j%5E2%2B3j-28=0 the discriminant is 9-%284%2A-4%2A-28%29 which is 9-%28448%29=+-439 so no real solutions.
The graph of -4j%5E2%2B3j-28=0 demonstrates this +graph%28+300%2C+200%2C+-50%2C+50%2C+-1000%2C+50%2C+-4x%5E2%2B3x-28%29+