Question 49776
{{{(sqrt (4x+1))+3=0)}}}
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{{{(sqrt (4x+1))+3-3=0-3)}}}
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{{{(sqrt (4x+1))=-3)}}}
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{{{(sqrt (4x+1))^2=(-3)^2}}}
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4x+1=9
4x+1-9=9-9
4x+1-9=0
4x-8=0
4x-8+8=0+8
4x/4=8/4
x=2
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check by plugging x=2 back into the original equation.  Remember that the square root is "plus" or "minus" a number, so try both.