Both the roots of the equation (x - b) (x - c) + (x - a) (x - c) + (x - a) (x - b) = 0 are always

If $$\ell $$, m, n are real, $$\ell \ne m$$, then the roots by the equation :
$$(\ell - m)\,{x^2} - 5\,(\ell + m)\,x - 2\,(\ell - m) = 0$$ are

If x, y and z are real and different and $$\,u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - 2xy$$, then u is always.

Let a > 0, b > 0 and c > 0. Then the roots of the equation $$a{x^2} + bx + c = 0$$