Question 35866
x^2-16x+32=0


First add -32 to each side and leave a blank space after the x^2- 16x .

{{{x^2 - 16x + ______= -32 + _____}}}


You must take half of the -16 (which is -4) and square it (which would be 16).  Add +16 to each side of the equation on the blank spaces I provided you.

{{{x^2 - 16x + 16= -32 + 16}}}


Now, notice that the left side is a perfect square trinomial:

{{{(x-4)^2 = -16}}}


Now you can take the square root of each side:
{{{x-4 = 0 +- sqrt(-16)}}}


Remember that {{{sqrt(-16) = 4i}}}??

{{{x-4 = 0 +- 4i)}}}


Last add +4 to each side:

{{{x = 4 +- 4i }}}


R^2 at SCC