Question 541963
The quadratic formula is:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
I will make the replacement
{{{ sqrt(b^2  - 4a*c) = z }}}
Now I have
{{{ x = (-b + z) / (2*a) }}}
and
{{{ x = (-b - z) / (2*a) }}}
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If {{{z}}} is not zero and is a real number, I
get a different solution for each equation,
so 2 solutions.
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If {{{z = 0}}}, then there is only 1 solution
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Also, since {{{z}}} is a square root, it can be
{{{ sqrt(-1*n) }}}, which is {{{ i*sqrt(n) }}}
and there are 2 imaginary solutions.
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These possibilities depend on what
{{{ b^2 - 4a*c }}} is
{{{ b^2 - 4a*c }}} = 0 ( one solution )
{{{ b^2 - 4a*c }}} = (+) ( two real solutions )
{{{ b^2 - 4a*c }}} = (-) ( two imaginary solutions )
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In the case of  {{{ b^2 - 4a*c }}} = (+), if this is a perfect square,
like {{{ 64 }}} The 2 real solutions will be rational, which means
a fraction of some kind.
If it is not a perfect square, like {{{31}}}, the 2 real solutions
will be irrational, like {{{ 3.666666 }}}. . . forever
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Hope this isn't confusing.