document.write( "Question 94385: solve the equation by using the quadratic formula\r
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Algebra.Com's Answer #68733 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Let's use the quadratic formula to solve for z:\r
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\n" ); document.write( "\n" ); document.write( "Starting with the general quadratic\r
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\n" ); document.write( "\n" ); document.write( "\"az%5E2%2Bbz%2Bc=0\"\r
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\n" ); document.write( "\n" ); document.write( "the general solution using the quadratic equation is:\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%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( "So lets solve \"3%2Az%5E2-8%2Az%2B2=0\" ( notice \"a=3\", \"b=-8\", and \"c=2\")\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%28--8+%2B-+sqrt%28+%28-8%29%5E2-4%2A3%2A2+%29%29%2F%282%2A3%29\" Plug in a=3, b=-8, and c=2\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+sqrt%28+%28-8%29%5E2-4%2A3%2A2+%29%29%2F%282%2A3%29\" Negate -8 to get 8\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+sqrt%28+64-4%2A3%2A2+%29%29%2F%282%2A3%29\" Square -8 to get 64 (note: remember when you square -8, you must square the negative as well. This is because \"%28-8%29%5E2=-8%2A-8=64\".)\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+sqrt%28+64%2B-24+%29%29%2F%282%2A3%29\" Multiply \"-4%2A2%2A3\" to get \"-24\"\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+sqrt%28+40+%29%29%2F%282%2A3%29\" Combine like terms in the radicand (everything under the square root)\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+2%2Asqrt%2810%29%29%2F%282%2A3%29\" Simplify the square root (note: If you need help with simplifying the square root, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+2%2Asqrt%2810%29%29%2F6\" Multiply 2 and 3 to get 6\r
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\n" ); document.write( "\n" ); document.write( "So now the expression breaks down into two parts\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B+2%2Asqrt%2810%29%29%2F6\" or \"z+=+%288+-+2%2Asqrt%2810%29%29%2F6\"\r
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\n" ); document.write( "\n" ); document.write( "Now break up the fraction\r
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\n" ); document.write( "\n" ); document.write( "\"z=%2B8%2F6%2B2%2Asqrt%2810%29%2F6\" or \"z=%2B8%2F6-2%2Asqrt%2810%29%2F6\"\r
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\n" ); document.write( "\n" ); document.write( "\"z=4+%2F+3%2Bsqrt%2810%29%2F3\" or \"z=4+%2F+3-sqrt%2810%29%2F3\"\r
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\n" ); document.write( "\n" ); document.write( "So these expressions approximate to\r
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\n" ); document.write( "\n" ); document.write( "\"z=2.38742588672279\" or \"z=0.279240779943873\"\r
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\n" ); document.write( "\n" ); document.write( "So our solutions are:\r
\n" ); document.write( "\n" ); document.write( "\"z=2.38742588672279\" or \"z=0.279240779943873\"\r
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\n" ); document.write( "\n" ); document.write( "Notice when we graph \"3%2Ax%5E2-8%2Ax%2B2\" (just replace z with x), we get:\r
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\n" ); document.write( "\n" ); document.write( "when we use the root finder feature on a calculator, we find that \"x=2.38742588672279\" and \"x=0.279240779943873\".So this verifies our answer
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