Question 682242
{{{(1/2)(4) = 2}}} and {{{2^2 = 4}}}
...
{{{x^2 + 4x + 4 = 12 + 4}}}
{{{(x+2)^2 = 16}}}
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{{{sqrt(x+2)^2}}} = +/- {{{sqrt(16)}}}
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x+2 = +/- 4
x = -2 +/- 4
...
x = -2+4 = {{{highlight_green(2)}}}
AND
x = -2-4 = {{{highlight_green(-6)}}}
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