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Solve x2 – 4x = 12 by completing the square

Mathematics
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\[\large{x^2-4x-12=0}\] right?
you just add 4 to both sides. that will make it \[(x-2)^{2}=16\]

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Other answers:

ok so now .. \[\large{x^2-4x-12=x^2-4x+4-16=0}\] right?
Yeah
now can u write \[\large{x^2-4x+4=(x-2)^2}\] ??
(x)^2 -2(2)(x)+2^2 = a^2-2ab+b^2=(a-b)^2=(x-2)^2
Stuck
ok so : (x^2-4x+4) = (x-2)^2 ?
x^2-4x-12=0 x=-2 x=6? or am I doing it wrong
I think you are doing it wrong ; : x^2-4x-12 = 0 \[\large{x^2-4x+4-16=0}\] \[\large{(x-2)^2-16=0}\] \[\large{(x-2-4)(x-2+4)=0}\] \[\large{(x-6)(x+2)=0}\] hence either x = 6 or x = -2
always remember that : \[\large{(a-b)^2=a^2-2ab+b^2}\] \[\large{(a+b)^2=a^2+2ab+b^2}\]
or : \[\large{(a\pm b)^2=a^2\pm 2ab +b^2}\]
You're awesome! Thank you so much! The formulas were very helpful :)
nice to hear .. :) best of luck

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