SOLUTION: How would I factor {{{ 5x^2+5x=20 }}} I was thinking that I would move the 20 to the other side of the equal sign so that it would look like {{{ 5x^2+5x-20=0 }}} and then factor

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: How would I factor {{{ 5x^2+5x=20 }}} I was thinking that I would move the 20 to the other side of the equal sign so that it would look like {{{ 5x^2+5x-20=0 }}} and then factor      Log On


   

Question 164492: How would I factor +5x%5E2%2B5x=20+ I was thinking that I would move the 20 to the other side of the equal sign so that it would look like +5x%5E2%2B5x-20=0+ and then factor it from there, but even then I'm still stuck on the factoring. Please help thank you.
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
+5x%5E2%2B5x=20+
+5x%5E2%2B5x-20=0+ So far, so good.
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Factor out a 5:
+5%28x%5E2%2Bx-4%29=0+
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That's as far as you can go. To solve you would need the apply the "quadratic equation".
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Doing so, would yield:
x = {1.562, -2.562}
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-4+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-4=17.

Discriminant d=17 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+17+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+17+%29%29%2F2%5C1+=+1.56155281280883
x%5B2%5D+=+%28-%281%29-sqrt%28+17+%29%29%2F2%5C1+=+-2.56155281280883

Quadratic expression 1x%5E2%2B1x%2B-4 can be factored:
1x%5E2%2B1x%2B-4+=+1%28x-1.56155281280883%29%2A%28x--2.56155281280883%29
Again, the answer is: 1.56155281280883, -2.56155281280883. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-4+%29