Complex Solutions to a Quadratic Equation 
Given equation:
  $$x^2 + 5x + c = 0$$
  For what $c$'s will the equation have non-real solutions?

I have try to plug in all kinds of different values for $c$ and then solve the equation to see if it gives me complex solution.
My question: Is there a theorem/lemma or a formula I need to state to answer this question?
 A: You have exactly one root if you have a perfect square.
Compete the square:
$(x^2 + 2 \frac 52 x + \frac {25}{4}) = (x+\frac 52)^2$
If $c = \frac {25}{4}$ there is one (real) root.
If $c >\frac {25}{4}$ your curve never touches the x axis. And you have no real roots.  If a quadratic has no real roots, it has complex roots.
if $c > \frac {25}{4}$ the curve crosses the x axis.  And it crosses on both sides of the vertex.
Alternate:  apply the quadratic formula
If you know the quadratic formula, then what does it take to get a negative under the radical?
A: Hint: This question is equivalent to finding the values of $c$ when $$b^2-4ac=5^2-4c<0$$
A: The discriminant formula for any quadratic is-
$D=b^2-4ac$, where $a,b$ and $c$ are the respective coefficients of the quadratic.


*

*If $D>0$, then the quadratic has two real and distinct roots. 

*If $D=0$, then the quadratic has two real and equal roots.

*If $D<0$, then the quadratic has two complex roots.



In your question-
$$a=1, b=5, c=?$$
$$D=5^2 - 4×1×c =25 - 4c$$
The equation has non-real roots only if-
$D<0$
$\Rightarrow 25 - 4c <0$
$\Rightarrow c < 25/4$ $_{(Answer)}$
