document.write( "Question 147093: 2x^2-4x-7=0 solve by completing the square \n" ); document.write( "
Algebra.Com's Answer #107427 by solver91311(24713)\"\" \"About 
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\"2x%5E2-4x-7=0+\"\r
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\n" ); document.write( "\n" ); document.write( "Step 1: Add the constant term to both sides\r
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\n" ); document.write( "\n" ); document.write( "\"2x%5E2-4x=7\"\r
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\n" ); document.write( "\n" ); document.write( "Step 2: Divide both sides by the lead coefficient\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-2x=7%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "Step 3: Divide the coefficient on the 1st degree term by 2 (\"%28-2%29%2F2\"), square the result \"%28-1%29%5E2=1\", then add it to both sides\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-2x%2B1=7%2F2%2B1\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-2x%2B1=9%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "Step 4: Factor the perfect square on the left\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-1%29%5E2=9%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "Step 5: Take the square root of both sides\r
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\n" ); document.write( "\n" ); document.write( "\"x-1=sqrt%289%2F2%29\" or \"x-1=-sqrt%289%2F2%29\"\r
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\n" ); document.write( "\n" ); document.write( "Step 6: Add the constant term on the left to both sides\r
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\n" ); document.write( "\n" ); document.write( "\"x=1%2B-sqrt%289%2F2%29\"\r
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\n" ); document.write( "\n" ); document.write( "Step 7: Simplify\r
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\n" ); document.write( "\n" ); document.write( "(Left as an exercise for the student. Remember to rationalize your denominator)
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