On some manipulations of $E[( X - tY)^2 ]$ I am following some notes and don't fully understand a step:
call $0 \le f(t) = E[ (X- tY)^2] $ then by linearity of the expected value we get
$$f(t) =E[Y^2]t^2 - 2 E[XY]t + E[X^2]$$
As noted this polynomial in $t$ has either two imaginary roots, or a real one when we have two equal solutions.
This gives us the inequality $E[XY]^2 \le E[X^2] E[Y^2]$ and also if I have $E[XY]^2 = E[X^2] E[Y^2]$ (the delta equal to zero) then:
(And here is the step I don't fully understand) this means that $X - t^* Y = c$ .
But shouldn't this mean that there exists a $t^*$ s.t. $E[ (X- t^*Y)^2] = a \in R$ 
I am not sure about the $X - t^* Y = c$ could someone explain it to me?
 A: Random variables are vectors: you can add them and multiply them by real numbers.  The recipe that takes a vector $X$ and returns the number $\mathbb{E}(X^2)$ defines a squared Euclidean length (if we do not distinguish two random variables that are equal almost surely). Similarly the recipe that takes two vectors $X,Y$ and returns the number $\mathbb{E}(XY)$ is the associated dot product. Let us therefore use vector notation where
$$|X|_2 = \sqrt{\mathbb{E}(X^2)}$$
and
$$X\cdot Y = \mathbb{E}(XY).$$
Since only two vectors are involved in this question, we are merely doing Euclidean geometry in a space of at most two dimensions: the Euclidean plane.  This should be easy!
Your notes, interpreted in this way, go as follows:
Let $f(t) = |X - tY|^2$ be the squared length of the vector $X-tY$. From
$$|X-tY|^2 = (X-tY)\cdot(X-tY) = |X|^2 - 2tX\cdot Y + t^2|Y|^2\tag{1}$$
and the fact that squares are nonnegative we deduce
$$0 \le |X|^2 - 2tX\cdot Y + t^2|Y|^2$$
for all $t$.  
Geometrically, $f(t)$ tracks the squared distance to $0$ along the line $t\to X - tY$ passing through $X$ in the $Y$ direction.  The distance itself is always nonnegative.  If it can ever become zero, that means this line passes through the origin.  Otherwise, the line does not intersect the origin.
The right hand side of $(1)$ is a polynomial in $t$ with coefficients $|Y|^2, -2X\cdot Y,$ and $|X|^2$.  Notice that a real root of this polynomial is any $t$ for which $|X-tY|^2 = 0$, which means $X-tY$ is the zero vector, equivalent to saying $X=tY$ (that is, $X$ is a multiple of $Y$).  Obviously there is at most one such $t$, showing there can be at most one real root.
From elementary algebra we know that a quadratic has one real root if and only its discriminant is zero.  Thus, $X$ is a multiple of $Y$ if and only if
$$(X\cdot Y)^2 - |X|^2|Y|^2 = 0.$$
Taking square roots, if both $X$ and $Y$ are nonzero this says
$$\frac{X\cdot Y}{|X| |Y|}=\pm 1.$$
The left hand side is the cosine of the angle subtended by $X$ and $Y$.  They are parallel if and only if this cosine has unit length; that is, the angle is a multiple of a straight angle.
These results about parallelism in the plane are familiar and trivial!  Perhaps by seeing them in this form the manipulations with random variables will become equally familiar, obvious, and memorable.

I cannot explain the business about "delta", $t^{*}$, $a$, or $c$, but I believe that the foregoing contains all the relevant points your notes were trying to get across.
