Question 685998
Solve for x by completing the square
{{{2x^2+20=4x}}}
First get the x's together and the constant on the other side.
{{{2x^2-4x=-20}}}
Then divide the whole equation by any coefficient in front of the x^2...in this case, 2.
{{{2x^2/2-4x/2=-20/2}}}
{{{x^2-2x=-10}}}
At this point, if we call the coefficient of x, b....then we would add (b/2)^2 to both sides.
{{{(b/2)^2=(-2/2)^2}}} -> {{{(-1)^2=1}}}
{{{x^2-2x+1=-10+1}}}
Factor the left and simplify the right.  Perfect squares always factor to be the sqrt or the front and back with the sign that is in front of the x term between them...like this.
{{{(x-1)^2=-9}}}
Take the square root of both sides...don't forget it is always + or -.
{{{sqrt((x-1)^2)}}}=+ or - {{{sqrt(-9)}}}
If you are only allowed to use "real" solutions...you are done...there is no "real" solution.  However, if you are using imaginary or complex numbers...keep going as follows:
x-1= + or -3i
x=1 + or -3i
This is two answers:
{{{highlight(x=1-3i)}}} or {{{highlight(x=1+3i)}}}
Happy Calculating!!!!