<s>[INST] Probability for roots of quadratic equation to be real, with coefficients being dice rolls. I really need help with this question.

The coefficients $a,b,c$ of the quadratic equation $ax^2+bx+c=0$ are determined by throwing $3$ dice and reading off the value shown on the uppermost face of each die, so that the first die gives $a$, the second $b$ and and third $c$. Find the probabilities that the roots the equations are real, complex and equal.

I was thinking about using the fundamental formula but i'm not sure how to go about doing it. Help would be greatly appreciated.
 [/INST] For roots to be real,
$ b^2 - 4ac >= 0 $
the following values of {b,a,c} are possible,
{2,1,1} 
{3,1,1} {3,1,2} {3,2,1} 
{4,1,1} {4,1,2} {4,2,1} {4,2,2} {4,3,1} {4,13,}
{5,1,1} {5,1,2} {5,2,1} {5,2,2} {5,3,2} {5,2,3} {5,3,1} {5,1,3}
{6,1,1} {6,1,2} {6,2,1} {6,2,2} {6,3,1} {6,1,3} {6,3,3} {6,3,2} {6,2,3}
These are 27 cases.
While the total cases possible for the values of a, b, c are 6*6*6
hence probability equals $\frac{27}{216}$ = $0.125$
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For the roots to be complex, 60 - 27 = 33 cases are possible,
probability equals $\frac{3}{216}$ = $0.152$
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For roots to be equal, discriminant 
$ b^2 - 4ac = 0 $
which is possible for the following values of {b,a,c} only {6,3,3}, {4,2,2}, {2,1,1},
probability equals  $\frac{3}{216}$ = $0.0138$
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