Proving the contradiction/negation of a statement Is proving the contradiction and proving the negation the same thing? Proving it true is proving the original statement false and proving it's false proves the statement is true?
So if a statement is an implication we just assume that the hypothesis is false? 
What if there's a quantifier in front? 
What would proving the negation of this statement look like?:
$\forall x \in \Bbb R$, if $x > 2$, then $x^2 + 3 > 0$?
 A: What you are actually asking about, according to the comments, is “proof by contradiction” (which is not “prove the contradiction”).
A proof by contradiction is a method of proof in which one assumes the negation of what you want to prove, and deduce a statement that is impossible. In classical logic, this means that the original statement must be true, because of the law of the excluded middle: it cannot be false (because if it were false that would lead to a contradiction), and if it is cannot be false, then it must be true. 
In order to do a proof by contradiction, you must know the negation of the statement; but you are not trying to prove that the negation is true. You are assuming that the negation is true, and trying to deduce a statement known to be false/impossible. 
In the case you give, the statement you want to prove is
$$\forall x\in\mathbb{R}\Bigl( x\gt 2 \rightarrow x^2+3\gt 0\Bigr)$$
The negation of this statement is
$$\begin{align*}
&\neg\Biggl( \forall x\in\mathbb{R}\Bigl(x\gt 2 \rightarrow x^2+3\gt 0\Bigr)\Biggr)\\
&\exists x\in\mathbb{R}\Biggl(\neg\Bigl( x\gt 2\rightarrow x^2+3\gt 0\Bigr)\Biggr)\\
&\exists x\in\mathbb{R}\Biggl( x\gt 2 \text{ and } \neg(x^2+3\gt 0)\Biggr)\\
&\exists x\in\mathbb{R}\Bigl( x\gt 2 \text{ and }x^2+3\leq 0\Bigr)
\end{align*}
$$
So, to do a proof by contradiction, you would start by assuming that there is a real number $x$ that is both greater than 2, and also has the property that $x^2+3\leq 0$. From this, you would want to deduce something utterly impossible.
If you were trying to prove the negation, you would be trying to prove that there is a real number $x$ that is both greater than 2 and also has the property that $x^2+3\leq 0$. 
Different things entirely. 
If you do a proof by contradiction successfully, you have established that the given statement is true.
If you successfully prove the negation, you have established that the given statement is false.
