Since you were given the last term, you can also apply the sum formula
$ \ s \ = \ \frac{n}{2}·(a_1 + a_n) \ \ , $ which produces $ \ 426 \ = \ \frac{n}{2}·(a_1 + 63) \ \Rightarrow \ ( a_1 + 63 ) · n \ = \ 852 \ \ . $ Putting this together with the equation you found, you can substitute either
$$ a_1 \ \ = \ \ 68 \ - \ 5n \ \ \Rightarrow \ \ ( 131 \ - \ 5n ) · n \ = \ 852 \ \ \Rightarrow \ \ 5n^2 \ - \ 131n \ + \ 852 \ \ = \ \ 0 $$
or
$$ n \ \ = \ \ \frac{68 \ - \ a_1}{5} \ \ \Rightarrow \ \ ( 63 \ + \ a_1 ) · \left( \frac{68 \ - \ a_1}{5} \right) \ = \ 852 \ \ \Rightarrow \ \ a_1^2 \ - \ 5a_1 \ - \ 24 \ \ = \ \ 0 \ \ . $$
Either choice leads to a quadratic equation. The latter is easier to factor, yielding $ \ (a_1 - 8)·(a_1 + 3) \ = \ 0 \ \ , $ from which we obtain
$$ \mathbf{a_1 \ = \ 8 \ \ : \ } \ \ n \ \ = \ \ \frac{68 \ - \ 8}{5} \ \ = \ \ 12 \ \ \ \text{or} \ \ \ \mathbf{a_1 \ = \ -3 \ \ : \ } \ \ n \ \ = \ \ \frac{68 \ - \ [-3]}{5} \ \ = \ \ 14.2 \ \ , $$
so only the first result is admissible. Checking the sum by the usual formula, we obtain $ \ 12·8 \ + \ \frac{12·11}{2}·5 \ = \ 96 + 330 \ = \ 426 \ \ . $ The first quadratic equation has a discriminant of $ \ 121 \ \ , $ so we have $ \ n \ = \ \frac{131 \ \pm \ 11}{2·5} \ = \ \frac{120}{10} \ = \ 12 \ \ , \ \ \frac{142}{10} \ = \ 14.2 \ \ ; $ the integer number of terms tells us that $ \ a_1 \ \ = \ \ 68 \ - \ 5·12 \ = \ 8 \ \ , $ which we have found to produce the correct sum.