Hello, MK!

These are of the same type . . . I'll explain the first one.

x² + x .= .0 .(mod 5)

We can apply standard algebraic methods.

Factor: .x(x + 1) .= .0 .(mod 5)

Then: .x .= .0 .(mod 5) . → . x .= .5m, for some integerm.

And: .x+1 .= .0 .(mod 5) . → . x .= .5n - 1, for some integern.