SOLUTION: Find all real or imaginary solutions to each equation. Use the method of your choice. 3y^2 + 4y

SOLUTION: Find all real or imaginary solutions to each equation. Use the method of your choice. 3y^2 + 4y - 1 =0 thank you

Question 177225: Find all real or imaginary solutions to each equation.
Use the method of your choice.
3y^2 + 4y - 1 =0
thank you
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
3y^2 + 4y - 1 =0
.
Your first attempt should be to try and factor the expression. In this case, you can't factor. Your next recourse is to use the quadratic equation. Doing so will yield the following two real solutions:
x = {0.215, -1.549}
.
Details of the quadratic follows:
.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=28 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.21525043702153, -1.54858377035486. Here's your graph:

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