Basically a basic problem

We know,

Area of triangle $(A) = \frac{1}{2}.BC.AB. \sin B$

Therefore,

$35=\frac{1}{2}(x-1)(x+3) \sin30°$

$(x-1)(x+3) = 35×2×2$

$x^{2}+2x-3 = 140$

$x^{2}+2x-143 = 0$

Now , its roots are ,

$x = \frac{-b± \sqrt{b^{2}-4ac}}{2a}$

$x = \frac{-2± \sqrt{2^{2}-4(1)(-143)}}{2(1)}$

$x = \frac{-2± \sqrt{4+572}}{2}$

$x = \frac{-2±24}{2}$

Thus , $x = 11,-13$ .

Here $x = -13$ is $Rejected$ because then the sides will be $Negative$ .

Therefore , $x = 11$ .

Thus , the $BC$ and $AB$ are $10$ and $14$ .

Sum of $AB+BC$ , $= 10+14 = \boxed{24}$