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 $x$

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

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

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

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

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