document.write( "Question 54275This question is from textbook Algebra for College Students
\n" ); document.write( ": I have the whole quadratic equation, just want to make sure that I am plugging in the numbers correctly. if I have \"3x%5E2-6x%2B2=0\" then, in my quadratic eq a=3,b=6, and c=2, is that correct? From there I should just follow through with the eq ? I am getting -2 sqrt 3 and it just doesn't look right. \n" ); document.write( "
Algebra.Com's Answer #36533 by funmath(2933)\"\" \"About 
You can put this solution on YOUR website!
Standard Form of a quadratic equation is \"highlight%28ax%5E2%2Bbx%2Bc=0%29\"
\n" ); document.write( "\"3x%5E2-6x%2B2=0\"
\n" ); document.write( "a=3, b=-6, c=2 (You had left the - off of the b.)
\n" ); document.write( "The Quadratic Formula is:
\n" ); document.write( "\"highlight%28x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29%29\"
\n" ); document.write( "Plug your numbers into the formula:
\n" ); document.write( "\"x=%28-%28-6%29%2B-sqrt%28%28-6%29%5E2-4%283%29%282%29%29%29%2F%282%283%29%29\"
\n" ); document.write( "\"x=%286%2B-sqrt%2836-24%29%29%2F6\"
\n" ); document.write( "\"x=%286%2B-sqrt%2812%29%29%2F6\"
\n" ); document.write( "\"x=%286%2B-sqrt%284%29%2Asqrt%283%29%29%2F6\"We factor 12 into 4*3 because 4 is a perfect square. The square root of 4 is 2, but we can't take the square root of three.
\n" ); document.write( "\"x=%286%2B-2%2Asqrt%283%29%29%2F6\" I can't fix this glitch...+/-is supposed to be in front of the 2.
\n" ); document.write( "\"x=2%283%2B-sqrt%283%29%29%2F%282%2A3%29\"The 2'2 cancel because 2/2=1.
\n" ); document.write( "\"x=cross%282%29%283%2B-sqrt%283%29%29%2F%28cross%282%29%2A3%29\"
\n" ); document.write( "\"highlight%28x=%283%2B-sqrt%283%29%29%2F3%29\"
\n" ); document.write( "Happy Calculating!!!
\n" ); document.write( "
\n" );