document.write( "Question 916678: Solve 8x2 + 6x + 5 = 0. \n" ); document.write( "
Algebra.Com's Answer #556218 by josh_jordan(263)\"\" \"About 
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\"8x%5E2%2B6x%2B5=0\"\r
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\n" ); document.write( "\n" ); document.write( "You will use the quadratic formula to solve for x. Recall that the quadratic formula is as follows: \"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"\r
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\n" ); document.write( "\n" ); document.write( "\"x=%28-6%2B-sqrt%286%5E2-4%2A8%2A5%29%29%2F%282%2A8%29\"----->\r
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\n" ); document.write( "\n" ); document.write( "\"x=%28-6%2B-sqrt%2836-160%29%29%2F%2816%29\"----->\r
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\n" ); document.write( "\n" ); document.write( "\"x=%28-6%2B-sqrt%28-124%29%29%2F%2816%29\"\r
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\n" ); document.write( "\n" ); document.write( "At this point, we will be using imaginary numbers, since the number inside the square root is negative. When we factor 124, we see that 4 x 31 = 124. We can then reduce the root to \"2i%2Asqrt%2831%29\". We now have\r
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\n" ); document.write( "\n" ); document.write( "\"x=%286%2B-2i%2Asqrt%2831%29%29%2F%2816%29\" (Ignore the ( between the +- symbol. It automatically formatted that way)\r
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\n" ); document.write( "\n" ); document.write( "Next, we can factor out a 2 from the numerator, reducing our numerator and denominator to the following, which is our final answer:\r
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\n" ); document.write( "\n" ); document.write( "\"x=%28-3%2Bi%2Asqrt%2831%29%29%2F%288%29\",\"%28-3-i%2Asqrt%2831%29%29%2F%288%29\"\r
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