What is a if the equation has real roots? ax^2+(3a-1)x+a+3=0

For the given equation to have real roots, the
discriminant delta has to be positive or zero.

delta =
(3a-1)^2 - 4a(a+3)

delta >=
0

We'll expand the square and we'll remove the brackets in
the expression of delta:

9a^2 - 6a + 1 - 4a^2 - 12a
>= 0

We'll combine like
terms:

5a^2 - 18a + 1 >=
0

We'll have to determine first the roots of the expression
5a^2 - 18a + 1:

5a^2 - 18a + 1 =
0

We'll apply the quadratic
formula:

a1 = [18+sqrt(324 -
20)]/10

a1 =
(18+4sqrt19)/10

a1 =
(9+2sqrt19)/5

a2 =
(9-2sqrt19)/5

The expression 5a^2 - 18a + 1
>= 0 for the values of a from the
intervals:

(-infinite ;
(9-2sqrt19)/5] U [(9+2sqrt19)/5 ; +infinite)