Question 419231
{{{ x^2 + 2x = 2 }}}
Take 1/2 of the coefficient of {{{x}}}, square it,
and add it to both sides
{{{x^2 + 2x + (2/2)^2 = 2 + (2/2)^2 }}}
{{{x^2 + 2x + 1 = 2 + 1 }}}
The left side is now a perfect square
{{{ (x + 1)^2  = 3 }}}
Now take the square root of both sides
{{{ x + 1 = sqrt(3) }}}
{{{x = -1 + sqrt(3) }}} answer
and
{{{ x + 1 = - sqrt(3) }}}
{{{x = -1 - sqrt(3) }}} answer
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These 2 answers are the roots. You can check 
your work this way:
{{{ x = r1}}}
{{{x = r2}}}
so,
{{{x - r1 = 0}}}
{{{x - r2 = 0}}}
so,
{{{x - (-1 + sqrt(3))  = 0}}}
{{{x - (-1 - sqrt(3))  = 0}}}
so,
{{{ (x + 1 - sqrt(3)) * (x + 1 + sqrt(3)) = 0}}}
Multiplying this out,
{{{ x^2 + x - sqrt(3)*x + x + 1 - sqrt(3) + sqrt(3)*x + sqrt(3) - 3 = 0 }}}
{{{ x^2 + 2x - 2 = 0 }}}
{{{ x^2 + 2x = 2 }}}
OK