Question 497544
{{{ x^2 - 2x + 10 = 0 }}}
When a 2nd degree equation is fairly
simple and the coefficient of the {{{ x^2 }}}
term is {{{ 1 }}}, as in your problem, the 
easiest approach is to complete the square.
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Do this by taking 1/2 of the coefficient of 
the {{{x}}} term, square it, then add it to
both sides, like this:
{{{ x^2 - 2x + 10 + (-2/2)^2 = 0 + (-2/2)^2 }}}
Now subtract {{{10}}} from both sides
{{{ x^2 - 2x + 1 = 1 - 10 }}}
The left side is now a perfect square.
The right side is a perfect square times {{{-1}}}
{{{ (x-1)^2 = -9 }}}
Take the square root of both sides
Note that {{{ sqrt(-1) = i }}}
{{{ x - 1 = 3*sqrt(-1) }}}
{{{ x - 1 = 3i }}}
{{{ x = 1 + 3i }}}
and also, since {{{ sqt(-9) }}} has a + and - answer,
{{{ x = 1 - 3i }}}
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Here's a plot of the equation. Notice that it never
touches the x-axis, which means the roots are
imaginary
{{{ graph( 400, 400, -10, 10, -2, 30, x^2 - 2x + 10) }}}