solve the quadratic equation by comleting the square, x^2-12x+7=0

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solve the quadratic equation by comleting the square, x^2-12x+7=0

Mathematics
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Sum 29 on both sides and you get:\[x^2-12x+36=(x-6)^2=29\]Do square root, isolate x and you get your answer.
huh?
Where did you get 29 from?

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(a+b)^2=a^2+2ab+b^2 To complete the squares is to go from something like the right part to something on the left side. a in this case is clearly x, and because of the second term, we know that b is 6, but we only have 7 on the place of b² and we need to have 36. What is missing for we to get 36 is what we need to add, wich is 29.
k i know im late on this but ill explain it anyway... \[x^2-12x+7=0\]we're gonna complete the square for this first move the \(7\) over to the other side:\[x^2-12x=-7\]now fill in the following blanks with the square of half the middle term [\(({12\over2})^2\)]:\[x^2-12x+\text{___}=-7+\text{___}\]\[x^2-12x+36=-7+36\]now combine like terms:\[x^2-12x+36=29\]next write the simplified version of \(x^2-12x+36\) which is \((x-6)(x-6)\)=\((x-6)^2\):\[(x-6)^2=29\]square root both sides:\[x-6=\sqrt{29}\]add 6 to both sides:\[\large x=6\pm\sqrt{29}\] hope you get how to do this now! :) ...sorry it was so late

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