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When the sum of a number plus 3 is squared, it is 11 more than the sum of the number plus 2 when squared.
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sum of a number plus 3 \(x+3\) squared \[(x+3)^2\] the sum of the number plus 2 squared \[(x+2)^2\] first one is eleven more than the second \[(x+3)^2=(x+2)^2+11\] solve for \(x\)
make sure to square correctly i.e. \((x+3)^2=x^2+6x+9\) and \((x+2)^2=x^2+4x+4\)
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