Paraphrasing Groucho Marx: if you don't like my first answer..., well I have another one. :-)
Here it is: let $X$ be a simply connected differentiable manifold.
Rational homotopy theory tells us that the rational homotopy type of $X$ (that is, its homotopy type modulo torsion) is contained in its minimal model, $M_X$, which is a commutative differential graded (cdg) algebra.
By definition, this means that you have a quasi-isomorphism (quis, a morphism of cdg algebras inducing an isomorphism in cohomology)
$$ M_X \longrightarrow \Omega^*(X) \ . $$
Here, $\Omega^* (X)$ is the algebra of differential forms of $X$ and the minimality of $M_X$ means that, in a certain, but precise, sense, it is the smaller cdg algebra for which such a quis exists.
The fact that $M_X$ contains the rational homotopy type of $X$ implies, for instance, that you can obtain the rangs of the homotopy groups of $X$ from it:
rang $\pi_n(X) =$ number of degree n generators (as an algebra) of $M_X$, for $n \geq 2$.
Nice, isn't it? :-)
The problem is that the algebra $\Omega^*(X)$ is, in general, not computable, so you can not obtain from it the minimal model $M_X$. And here is where formality comes to help you.
Almost by definition, $X$ is a formal space if there exists two quis
$$ \Omega^*(X) \longleftarrow M_X \longrightarrow H^*(X;\mathbb{Q}) $$
Hence, if $X$ is formal you can compute its minimal model $M_X$, and hence its rational homotopy type, directly from the cohomology algebra $H^*(X; \mathbb{Q})$, which is nicer (smaller, more computable) than $\Omega^*(X)$.
And the final point is that there are plenty of examples of spaces which are known to be formal.
(Final remark: Actually, you'd have to put $A_{PL}^*(X;\mathbb{Q})$ instead of $\Omega^*(X)$ to work over the rationals, but this you can find it explained in the references we have provided for you.)