To solve the first equation, expand the left side:

(x-5)(x-7) = x^2 - 7x - 5x + 35 = x^2 - 12x + 35

Setting this equal to 0:

x^2 - 12x + 35 = 0

This can be factored as:

(x-5)(x-7) = 0

So the solutions are x=5 and x=7.

For the second equation:

3x(x-4) = 0

This equation is already factored. The solutions are x=0 and x=4.