If $G$ is solvable and $H\trianglelefteq G$ then $G/H$ is solvable? I'd like to show the following result of group theory:

Problem: Let $G$ be a solvable group and $H\trianglelefteq G$. Then $G/H$ is solvable.

Definition: A group $G$ is said to be solvable if there is a normal series of $G$ such that the factors are abelian, that is, there exists a sequence of normal subgroups:$$G=G_0\trianglerighteq G_1\trianglerighteq \ldots \trianglerighteq G_n=\{1\},$$ such that the factors $G_{i}/G_{i+1}$ are abelians.

Sketch: Let $$G=G_0\trianglerighteq G_1\trianglerighteq \ldots \trianglerighteq G_n=\{1\},$$

a normal series of $G$ such that $G_i/G_{i+1}$ are abelians. The first try would be to consider 

$$\frac{G}{H}=\frac{G_0}{H}\trianglerighteq \frac{G_1}{H}\trianglerighteq \ldots \trianglerighteq \frac{G_n}{H}.$$

Of course this makes no sense for we don't even have $H\subset G_i$ for some $i$. But we can get subgroups of $G$ containing $H$ considering $HG_i$. Since $H\trianglelefteq G$ we have $HG_i\leq G$ for all $i$. So it is natural to consider the sequence: $$G=HG_0\geq HG_1\geq \ldots \geq HG_n=H\geq \{1\}.$$

It is not hard to verify this is a normal series of $G$. I conjecture the normal series for $G/H$ will arise from this but I still can't see how. 

Can anyone help me finishing this proof?
Consider the series
$$
G/H=(HG_0)/H \ge (HG_1)/H \ge \dots\ge (HG_n)/H=\{1\}.
$$
Homomorphic image of a solvable group is solvable.
Consider  natural homomorphism from G to G/H.