SOLUTION: 3x3x^2+7=0 3x^2+2x-7=0 x=-2+or-sqrt2^2-4(3)(-7)/2(3) x=-2+or-sqrt4+84/6 x=-2+or-sqrt88/6 I am stuck at this point do I break the radicand 4*22? Thank you, Tina

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 3x3x^2+7=0 3x^2+2x-7=0 x=-2+or-sqrt2^2-4(3)(-7)/2(3) x=-2+or-sqrt4+84/6 x=-2+or-sqrt88/6 I am stuck at this point do I break the radicand 4*22? Thank you, Tina      Log On


   

Question 237680: 3x3x^2+7=0

3x^2+2x-7=0

x=-2+or-sqrt2^2-4(3)(-7)/2(3)
x=-2+or-sqrt4+84/6
x=-2+or-sqrt88/6
I am stuck at this point do I break the radicand 4*22?

Thank you,

Tina
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You're ok except that:
sqrt%2888%29+=+%282%29sqrt%2822%29 so you can simplify the solutions a bit more:
x+=+%281%2F3%29%28-1%2Bsqrt%2822%29%29 or %281%2F3%29%28-1-sqrt%2822%29%29