With a standard form quadratic, we see that with the quadratic formula:
[tex] \frac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
That the part under the radical ([tex] b^2-4ac [/tex]) can tell us many things about how the roots of this polynomial behave.
If b^2-4ac were 0, that would mean the polynomial would have one root at [tex] \frac{-b}{2a} [/tex]. We call that a double root at (-b/2a,0).
If the discriminant is a perfect square, we note that the radical can be reduced and this the fraction is a rational number.
If the discriminant is negative, it means it has no real roots. That's because the root of a negative number is imaginary.
The last case if is the discriminant is neither a negative or a perfect square. That means the radical cannot be reduced and we will have two irrational roots.
In this problem, we see that the discriminant is:
[tex] 3^2-4*1(-4)= \\9+16=\\25 [/tex]
The answer to the problem you asked is then 25, and we can also note that this quadratic has real, rational roots.
