Question 36230: Good Morning: I haven't been able to solve this problem... I have to use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; just determine the number and types of solutions.
2x^2 - 6x + 5 = 0
Answer by Prithwis(166)   (Show Source):
You can put this solution on YOUR website! Discriminant = (b^2 - 4ac) for ax^2+bx+c
Here a = 2, b=-6 and c = 5
b^2 - 4ac = 36-40 = -4
If b^2 - 4ac =0, there is only one real solution.
If b^2 - 4ac > 0, both the roots are real.
If b^2 - 4ac < 0, both the roots are non-real complex.
Here the discriminant is -4 (<0); so there are two non-real complex solutions of the given equation.
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