Question 686643
When you have the equation in the form
{{{ y = a*x^2 + b*x + c }}}, the the discriminant is
{{{ b^2 - 4*a*c }}}
The rules are:
(1) If {{{ b^2 - 4*a*c < 0 }}}, then there are 2 imaginary roots
(2) If {{{ b^2 - 4*a*c > 0 }}}, then there are 2 real roots
(3) If {{{ b^2 - 4*a*c = 0 }}}, then there is a single real root ( called a double root )
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Your equation is:
{{{ f(r) = r^2 - (3/5)*r + 2/25 }}}
{{{ r^2 - (3/5)*r + 2/25 = 0 }}}
( note that this is saying {{{ f(r) = 0 }}}, which means that the
values of {{{r}}} that you find will be the x-crossings which are the roots )
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The fact that you have {{{r}}} and not {{{x}}} doesn't matter
The form is still {{{ a*x^2 + b*x + c = 0 }}}
{{{ a = 1 }}}
{{{ b = 3/5 }}}
{{{ c = 2/25 }}}
Now just plug these into the discriminant
{{{ b^2 - 4*a*c = (3/5)^2 - 4*1*(2/25) }}}
{{{ b^2 - 4*a*c = 9/25 - 8/25 }}}
{{{ b^2 - 4*a*c = 1/25 }}} 
This is a positive result, so there are 2 roots, and they are real
Now to solve equation:
{{{ r^2 - (3/5)*r + 2/25 = 0 }}}
{{{ ( r - 2/5 )*( r - 1/5 ) = 0 }}} ( I could see the result by inspection )
{{{ r = 2/5 }}} and
{{{ r = 1/5 }}} are the solutions
But here's solution by completing the square:
{{{ r^2 - (3/5)*r = -2/25 }}}
{{{ r^2 - (3/5)*r + (3/10)^2 = -2/25 + (3/10)^2 }}}
{{{ r^2 - (3/5)*r + 9/100 = -2/25 + 9/100 }}}
{{{ r^2 - (3/5)*r + 9/100 = -8/100 + 9/100 }}}
{{{ r^2 - (3/5)*r + 9/100 = 1/100 }}}
{{{ ( r - 3/10 )^2 = (1/10)^2 }}}
take the square root of both sides
{{{ r - 3/10 = 1/10 }}}
{{{ r = 4/10 }}}
{{{ r = 2/5 }}}
But, also I can say that
{{{ r - 3/10 = -1/10 }}}
Taking the negative square root of (1/10)^2
{{{ r = 3/10 - 1/10 }}}
{{{ r = 2/10 }}}
{{{ r = 1/5 }}}
I get the same answers
I hope this helps