## anonymous 4 years ago 3p^2+12p-2=0 how do you solve this problem? I need all the steps please and thank you.

1. anonymous

You have to use the Quadradic Formula here, it's not solvable by factoring: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

2. anonymous

To be clear that's for the standard form: ax^2 + bx + c = 0 a = 3 b = 12 c = -2 See how that works? Yes, you will get two answers, as you should (because of the x$$^2$$)

3. anonymous

okay

4. anonymous

If what is under the radical/square-root above is negative you'd have two imaginary roots, thankfully you won't here. But you will have square-roots (or fractional powers if you prefer) in your final answers.

5. anonymous

okay

6. anonymous

is it $x=12\pm \sqrt{120} \over 6$