How do you solve #x^2 - 3x + 6 = 0# by completing the square?

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Answer 1 · 2017-07-05T07:58:27

#x=3/2+-isqrt15/2#

#x^2-3x+6=0#

add and subtract half the coefficient of #x# squared

#color(blue)((x^2-3x+(3/2)^2))+6- ( 3/2)^2=0#

the section in blue is a perfect square

#color(blue)((x-3/2)^2)+15/4=0#

solve by rearranging for #x#

#(x-3/2)^2=-15/4#

#x-3/2=sqrt(-15/4)#

#x-3/2=+-isqrt15/2#

#x=3/2+-isqrt15/2#