SOLUTION: How many real number solutions are there to the equation 0=-3x^2+x-4?

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Question 948989: How many real number solutions are there to the equation 0=-3x^2+x-4?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

0=-3x%5E2%2Bx-4 ....use discriminant to determine what kind of solutions you have in this case
Evaluate b%5E2-4ac
If the value is b%5E2-4ac%3E0 there are two+real unequal roots
If the value is b%5E2-4ac=0 there is one real root, or two equal +real roots
If the value is b%5E2-4ac%3C0 there are two unequal complex roots
In this case you get b%5E2-4ac=1%5E2-4%2A%28-3%29%2A%28-4%29=1-4%2A12=1-48=-47
so, -47%3C0 and you have two unequal complex roots
answer to the question how many real number solutions are there is zero or none