SOLUTION: real or imaginary solutions 84. 3y^2 + 4y – 1 = 0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: real or imaginary solutions 84. 3y^2 + 4y – 1 = 0      Log On


   

Question 59849: real or imaginary solutions
84. 3y^2 + 4y – 1 = 0
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
real or imaginary solutions
84. 3y^2 + 4y – 1 = 0
We use the discriminant to determine if there are real or imaginary solutions. If the discriminant is negative there are two imaginary solutions. If it's positive there are two real solutions. If it's 0 there is one real solution.
The discriminant is highlight%28b%5E2-4ac%29
a=3, b=4, and c=-1
%284%29%5E2-4%283%29%28-1%29
16%2B12
28
It's positive so there's two real solutions. Because this is not a perfect square you have to use the quadratice formula if you have to solve it. (you could complete the square, but because a is 3, you would have to deal with fractions.)
Happy Calculating!!