quadratic

If we solve for x, x could most likely be 0:

x = 0

2x^2+3x+2=0

x Replaced with 0

2 * 0² + 3 * 0 + 2 = 0

I'm sticking with 0 as the answer.

If anyone has another answer, kudos to them for finding it. I can't find any others.

You can use the Quadratic Formula of 'completing the squares ' method to solve this

I'll use quadratic formula:\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

x = (-3 +- sq rt( 9-16))/4     =  -3/4 +- sqrt(-7/16)   =  - 3/4 +- sqrt(7/16) i