Question 948575
{{{m(x^2-x)+1-m=x^2}}}
Quadratic equation with "unknown" x, and constant m.
Real roots?  Discriminant must be non-negative.


{{{mx^2-mx+1-m-x^2=0}}}
{{{mx^2-x^2-mx+1-m=0}}}
{{{(m-1)x^2-mx+(1-m)=0}}}, the given equation now in general form.


Inequality for the discriminant needed:
{{{highlight_green(m^2-4(m-1)(1-m)>=0)}}}, left member is the discriminant.
{{{m^2-4(m-1)(-1)(m-1)>=0}}}
{{{m^2+4(m-1)^2>=0}}}
{{{m^2+4(m^2-2m+1)>=0}}}
{{{m^2+4m^2-8m+4>=0}}}
{{{highlight_green(5m^2-8m+4>=0)}}}


The truth of this m inequality depends on no real roots for m, meaning the discriminant of just the m inequality must be NEGATIVE.
If {{{(-8)^2-4*5(4)<0}}}, then the m quadratic inequality has no real roots.
{{{64-20*4<0}}}
{{{64-80<0}}}
{{{highlight_green(-16<0)}}}------TRUE.


That truth means that the {{{5m^2-8m+4>=0}}}  is satisfied; and that means that the original given equation will have real roots (for all m).