document.write( "Question 668534: What is the number of real solutions?
\n" ); document.write( "6x2 – 7x – 1 = 0 \n" ); document.write( "
Algebra.Com's Answer #415645 by swincher4391(1107)\"\" \"About 
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The easiest way to answer this question is to look at D, the discriminant. Now what is the discriminant? It may look familiar from your quadratic formula. The discriminant is \"b%5E2-4ac\". If:\r
\n" ); document.write( "\n" ); document.write( "b^2-4ac > 0 two real solutions
\n" ); document.write( "b^2-4ac = 0 one real solution
\n" ); document.write( "b^2-4ac < 0 no real solutions\r
\n" ); document.write( "\n" ); document.write( "So where a = 6 b = -7 and c = -1 we have that b^2 - 4ac = 49 - 4*6*(-1) = 49 + 24 = 63 > 0.\r
\n" ); document.write( "\n" ); document.write( "So, there are two real solutions.\r
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\n" ); document.write( "\n" ); document.write( "To see why b^2-4ac is the key, just think about your quadratic for a minute.\r
\n" ); document.write( "\n" ); document.write( "Since in our quadratic formula we have that \"%28-b+%2B-+sqrt%28b%5E2-4ac%29%29%2F%282a%29\" what happens when b^2-4ac > 0? Then we have some number that we are plus-or-minusing gives us two solutions. If b^2-4ac = 0, then plus-or-minus has no impact and in fact we get -b/2a which is our vertex's x-coordinate. If b^2-4ac is negative then we are taking the square root of a negative number and so now we have imaginary solutions -- none real. \n" ); document.write( "
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