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\[\large{x^2-4x-12=0}\] right?

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

@mathslover yes

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

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