For what values of $ a, b$ does the equation have real roots? 
For what values of $a,b$ does the equation
  $${ x }^{ 2 }+2\left( 1+a \right) x+\left( 3{ a }^{ 2 }+4ab+4{ b }^{ 2 }+2 \right) = 0$$
  have real roots?

For it to have real roots, the discriminant has to be $>0$, correct? (Or equal to, I suppose, since the question didn't specify distinct or not) So I tried using the values, which gave me ${ \left( 2+2a \right)  }^{ 2 }-4\left( 3{ a }^{ 2 }+4ab+4{ b }^{ 2 } +2\right) $ but I'm not sure where to go after that.
 A: You have the right idea.
You just need to continue
expanding that expression.
From
$x^2 + 2(1+a)x + (3a^2 + 4ab + 4b^2 + 2) = 0
$,
the discriminant is
(ignoring the factor of 2
since we are concerned
only about the sign)
$\begin{array}\\
d
&=(1+a)^2-(3a^2 + 4ab + 4b^2 + 2)\\
&=a^2+2a+1-(3a^2 + 4ab + 4b^2 + 2)\\
&=-2a^2+2a-1-4ab-4b^2\\
&=-a^2+2a-1-a^2-4ab-4b^2
\quad\text{This is the key step}\\
&=-(a-1)^2-(2b+a)^2\\
\end{array}
$
So,
in the miraculous way of many homework problems,
this is the negative
of a sum of two squares.
So $d \le 0$
and,
for $d = 0$,
we must have
$a=1$ and
$2b+a=0$,
which means
$b = -1/2$.
For any other values of
$a$ and $b$,
the discriminant is negative,
and so there are no real roots.
For these values of $a$ and $b$,
there is a repeated root.
A: Here it is easy to complete the square $$(x+1+a)^2+3a^2+4ab+4b^2+2-(1+a)^2=0$$ so that $$(x+1+a)^2=-2a^2+2a-1-4b^2-4ab=-(a-1)^2-(2b+a)^2$$
This step involves simply completing the square for $4b^2+4ab=(2b+a)^2-a^2$ and seeing what is left.
The left-hand side is non-negative, the right-hand side non-positive, so equality is only possible if both are zero.
Note: this is wholly equivalent to working with the discriminant, but saves a factor of $4$ and does not require remembering a formula.
