document.write( "Question 672676: I am stumped.. when I solved the equation there was no need for the square root method. But that is what my hmwrk question is calling for. Will you please break this problem down for me so that I can see where I am messing up?\r
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Algebra.Com's Answer #418256 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The other tutor's answer is only half right for there are TWO\r\n" );
document.write( "SOLUTIONS, not just one.  Here is the complete answer:\r\n" );
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document.write( "(x-3)² = 25\r\n" );
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document.write( "There are two possibilities for that since \r\n" );
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document.write( "(+5)² = 25 and (-5)² = 25 are BOTH TRUE.  Therefore,\r\n" );
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document.write( "(x-3) could either equal to +5 or (x-3) could equal to -5.\r\n" );
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document.write( "So we have two equations and therefore two solutions:\r\n" );
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document.write( "(x-3) = +5  and (x-3) = -5\r\n" );
document.write( "  x-3 = 5         x-3 = -5 \r\n" );
document.write( "    x = 8           x = -2\r\n" );
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document.write( "Edwin
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