What is true about the solutions of a quadratic equation when the radicand in the quadratic formula is negative?

Question

anonymous · Answer

No real solutions 
Two identical rational solutions
 Two different rational solutions
 Two irrational solutions

anonymous · Answer

i believe it is B

anonymous · Answer

LOL okay thx so much =)

anonymous · Answer

May i ask y you think that it's B?

anonymous · Answer

@mattsouth0015

anonymous · Answer

If we look at the quadratic formula:
$$
ax^2 + bx + c = 0 \
x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$Then the radicand will be the expression inside the radical which is $ (b^2 - 4ac) $
But if the radicant is negative then its square root $ \sqrt{b^2 - 4ac} $ cannot be a real number, because there is no real number that squares to be a negative number, so it will result in a complex number.
That will make the solutions for $x$ be complex as well and not real solutions. So the answer should be A

anonymous · Answer

Ugh! Oh well thx for the help @pitamar =)

anonymous · Answer

You're welcome =)