Question 269193
The discriminant is {{{b^2 - 4ac}}} in the formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
As long as {{{b^2 - 4ac }}} is (+), then the roots are real
You are given {{{s^2 - 5s + 6 = 0}}}
{{{b^2 - 4ac = (-5)^2 - 4*1*6}}}
{{{b^2 - 4ac = 25 - 24}}}
{{{b^2 - 4ac = 1}}}
There are 2 real roots, If the discrinant were {{{0}}},
there would be 1 real root