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(sqrtx-2)(-2)=x
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\[\sqrt{x-2}-2=x\]
If that's what it is, your notation is wrong in the original one
should've been sqrt(x-2) - 2 = x
\[\sqrt{x-2}=x \div(-2)\] x-2=x^2/4 4x-8=x^2 x^2-4x+8=0
i know thats why i rewrote it because i messed up, sorry :)
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Nick Black it's x - 2 not x/2
\[\sqrt{x-2}^{2}=(x+2)^{2}\] \[x-2=x ^{2}+2x+4\] -x^2 -x -6 = 0 x^2 +x+6 =0
it can't be finish with factor so you may use x=(-b+_b^2-4ac)/2a
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