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.
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.