Question 574827
The quadratic formula to find the roots of any
quadratic in the form {{{ f(x) = ax^2 + bx + c }}}
is  {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
Notice there is a (+) and (-) in front of the 
square root so you get 2 solutions, as you
should with a quadratic.
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The expression inside the square root is called
the discriminant. If the discriminant is a negative
number, you get a (+) and a (-)  imaginaries
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Here's the possibilities:
{{{ b^2 - 4a*c }}} = negative  > 2 imaginary roots
{{{ b^2 - 4a*c }}} = positive  > 2 real roots
{{{ b^2 - 4a*c }}} = 0  > 1 real root
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Here's your problem:
{{{ 3x^2 + x + 2 = 0 }}}
{{{ a = 3 }}}
{{{ b = 1 }}}
{{{ c = 2 }}}
{{{ b^2 - 4a*c = 1^2 - 4*3*2 }}}
{{{ b^2 - 4a*c = 1 - 24 }}}
{{{ b^2 - 4a*c = -23 }}}
This is an example of the 1st case, 2 imaginary roots
What this means is: the quadratic never touches the
x-axis, so there are no "real" roots.
Here's a plot to show this:
{{{ graph( 400, 400, -6, 6, -6, 6, 3x^2 + x + 2 ) }}}
Notice that if {{{a}}} or {{{c}}} were negative, you would
get real roots