document.write( "Question 84871This question is from textbook Elementary and intermediate Algebra
\n" ); document.write( ": Find all real or imaginary roots of 5x^2 + 5x + 4 = 0.
\n" ); document.write( "(If there is more than one root, show additional roots
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Algebra.Com's Answer #61144 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
We can find the roots of this quadratic with the quadratic formula\r
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"5x%5E2%2B5x%2B4+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%285%29%5E2-4%2A5%2A4=-55\".
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\n" ); document.write( " The discriminant -55 is less than zero. That means that there are no solutions among real numbers.

\n" ); document.write( " If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

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\n" ); document.write( " In the field of imaginary numbers, the square root of -55 is + or - \"sqrt%28+55%29+=+7.41619848709566\".
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\n" ); document.write( " The solution is
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\n" ); document.write( " Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5%2Ax%5E2%2B5%2Ax%2B4+%29\"

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\n" ); document.write( "\n" ); document.write( "So the quadratic has two imaginary roots\r
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\n" ); document.write( "\n" ); document.write( "\"x=5%2Bi%2Asqrt%2811%29%2F2\" or \"x=5-i%2Asqrt%2811%29%2F2\"
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