Subtracting lower-order term to prove subtitution method works
If you're using the same book I have (Introduction to Algorithms 3rd edition), then you mis-wrote the question. The relation in question should be: $T(n)=4T(n/2) + n.$

That said, substitution here won't work, as you know.  Assuming $T(n) \le cn^2,$
you get:

$T(n)\le4T(n/2) + n$

$T(n)\le4(c\frac{n^2}{4}) + n$

$T(n)\le cn^2 + n$

Which doesn't imply  $T(n) &lt; cn^2.$ Instead, you can make the assumption: $T(n) \le cn^2 - bn.$

$T(n)\le4T(n/2) + n$

$T(n)\le4(c\frac{n^2}{4}-b\frac{n}{2}) + n$

$T(n)\le cn^2 + n(-2b + 1)$

Letting $b=1,$ you get an equation consistent with the initial asumption. Namely:

$T(n)\le cn^2 - n$

As a side note, the relation you initially wrote, $T(n)=4T(n/2) + n^2,$ can be solved using case 2 of Master's Theorem, and has an answer of $\Theta(n^2lg(n))$