document.write( "Question 256556: Please find the quadratic formula of 6x^2+10x-4=0
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Algebra.Com's Answer #188655 by richwmiller(17219) $\"\"$ $\"About$

You can put this solution on YOUR website!
Do you mean solve using the quadratic formula?
\n" ); document.write( " 2(x+2)(3x-1) = 0
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

$\"6%2Ax%5E2%2B10%2Ax-4\"$ Start with the given expression.

$\"2%283x%5E2%2B5x-2%29\"$ Factor out the GCF $\"2\"$ .

Now let's try to factor the inner expression $\"3x%5E2%2B5x-2\"$

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Looking at the expression $\"3x%5E2%2B5x-2\"$ , we can see that the first coefficient is $\"3\"$ , the second coefficient is $\"5\"$ , and the last term is $\"-2\"$ .

Now multiply the first coefficient $\"3\"$ by the last term $\"-2\"$ to get $\"%283%29%28-2%29=-6\"$ .

Now the question is: what two whole numbers multiply to $\"-6\"$ (the previous product) and add to the second coefficient $\"5\"$ ?

To find these two numbers, we need to list all of the factors of $\"-6\"$ (the previous product).

Factors of $\"-6\"$ :

1,2,3,6

-1,-2,-3,-6

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to $\"-6\"$ .

1*(-6) = -6
2*(-3) = -6
(-1)*(6) = -6
(-2)*(3) = -6

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"5\"$ :

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First Number	Second Number	Sum
1	-6	1+(-6)=-5
2	-3	2+(-3)=-1
-1	6	-1+6=5
-2	3	-2+3=1

From the table, we can see that the two numbers $\"-1\"$ and $\"6\"$ add to $\"5\"$ (the middle coefficient).

So the two numbers $\"-1\"$ and $\"6\"$ both multiply to $\"-6\"$ and add to $\"5\"$

Now replace the middle term $\"5x\"$ with $\"-x%2B6x\"$ . Remember, $\"-1\"$ and $\"6\"$ add to $\"5\"$ . So this shows us that $\"-x%2B6x=5x\"$ .

$\"3x%5E2%2Bhighlight%28-x%2B6x%29-2\"$ Replace the second term $\"5x\"$ with $\"-x%2B6x\"$ .

$\"%283x%5E2-x%29%2B%286x-2%29\"$ Group the terms into two pairs.

$\"x%283x-1%29%2B%286x-2%29\"$ Factor out the GCF $\"x\"$ from the first group.

$\"x%283x-1%29%2B2%283x-1%29\"$ Factor out $\"2\"$ from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

$\"%28x%2B2%29%283x-1%29\"$ Combine like terms. Or factor out the common term $\"3x-1\"$

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So $\"2%283x%5E2%2B5x-2%29\"$ then factors further to $\"2%28x%2B2%29%283x-1%29\"$

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Answer:

So $\"6%2Ax%5E2%2B10%2Ax-4\"$ completely factors to $\"2%28x%2B2%29%283x-1%29\"$ .

In other words, $\"6%2Ax%5E2%2B10%2Ax-4=2%28x%2B2%29%283x-1%29\"$ .

Note: you can check the answer by expanding $\"2%28x%2B2%29%283x-1%29\"$ to get $\"6%2Ax%5E2%2B10%2Ax-4\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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\n" ); document.write( "\n" ); document.write( "the quadratic formula
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Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation $\"ax%5E2%2Bbx%2Bc=0\"$ (in our case $\"6x%5E2%2B10x%2B-4+=+0\"$ ) 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 $\"b%5E2-4ac\"$ : $\"b%5E2-4ac=%2810%29%5E2-4%2A6%2A-4=196\"$ .
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\n" ); document.write( " Discriminant d=196 is greater than zero. That means that there are two solutions: $\"+x%5B12%5D+=+%28-10%2B-sqrt%28+196+%29%29%2F2%5Ca\"$ .
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\n" ); document.write( " $\"x%5B1%5D+=+%28-%2810%29%2Bsqrt%28+196+%29%29%2F2%5C6+=+0.333333333333333\"$
\n" ); document.write( " $\"x%5B2%5D+=+%28-%2810%29-sqrt%28+196+%29%29%2F2%5C6+=+-2\"$
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\n" ); document.write( " Quadratic expression $\"6x%5E2%2B10x%2B-4\"$ can be factored:
\n" ); document.write( " $\"6x%5E2%2B10x%2B-4+=+6%28x-0.333333333333333%29%2A%28x--2%29\"$
\n" ); document.write( " Again, the answer is: 0.333333333333333, -2.\n" ); document.write( "Here's your graph:
\n" ); document.write( " $\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+6%2Ax%5E2%2B10%2Ax%2B-4+%29\"$

\n" ); document.write( "\n" ); document.write( " 2(x+2)(3x-1) = 0 \n" ); document.write( "

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