Question 171112
Complete the square on {{{2x^2-2x+7=0}}}


Step 1, starting from a quadratic in standard form like you have is to put the constant term on the right:


{{{2x^2-2x=-7}}}


Step 2, if the coefficient on the {{{x^2}}} term is other than 1, divide by that coefficient.


{{{x^2-x=-7/2}}}


Step 3, divide the resulting coefficient on the {{{x}}} term by 2 and square the result


{{{((-1)/2)^2=1/4}}}


Step 4, add this result to both sides of your equation and collect terms


{{{x^2-x+1/4=-7/2+1/4}}}


{{{x^2-x+1/4=-13/4}}}


Step 5, the above result has a perfect square on the left (hence the term "completing the square"), so factor it:


{{{(x-1/2)^2=-13/4}}}


Step 6, take the square root of both sides:


{{{x-1/2=sqrt(-13/4)}}} or {{{x-1/2=-sqrt(-13/4)}}}


Which leads us to a great big oops! because you can't take the square root of a negative number.  The solution is to use the imaginary number {{{i}}} which is defined as {{{i^2=-1}}}, leaving us with:


{{{x-1/2=sqrt((-1)13/4)}}} or {{{x-1/2=-sqrt((-1)13/4)}}}
{{{x-1/2=sqrt(-1)*sqrt(13/4)}}} or {{{x-1/2=-sqrt(-1)*sqrt(13/4)}}}
{{{x-1/2=i*sqrt(13/4)}}} or {{{x-1/2=-i*sqrt(13/4)}}}


Step 7, isolate {{{x}}} and simplify in each equation


{{{x[1]=1/2+i*sqrt(13/4)=(1+i*sqrt(13))/2}}}


{{{x[2]=1/2-i*sqrt(13/4)=(1-i*sqrt(13))/2}}}


If you want the exact representation of the roots of the given equation, you are done.  If you need a numerical approximation of the imaginary parts of your complex numbers, get out your calculator.


Multiplying {{{ (x-((1+i*sqrt(13))/2))(x-((1-i*sqrt(13))/2))}}} to verify that the product is, in fact, {{{2x^2-2x+7}}}, is left as an exercise for the student.  Alternatively, you could just trust me.