SOLUTION: solve for x x^2+12x+4=0

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Question 323630: solve for x
x^2+12x+4=0
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B12x%2B4+=+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=%2812%29%5E2-4%2A1%2A4=128.

Discriminant d=128 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-12%2B-sqrt%28+128+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2812%29%2Bsqrt%28+128+%29%29%2F2%5C1+=+-0.343145750507619
x%5B2%5D+=+%28-%2812%29-sqrt%28+128+%29%29%2F2%5C1+=+-11.6568542494924

Quadratic expression 1x%5E2%2B12x%2B4 can be factored:
1x%5E2%2B12x%2B4+=+1%28x--0.343145750507619%29%2A%28x--11.6568542494924%29
Again, the answer is: -0.343145750507619, -11.6568542494924. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B12%2Ax%2B4+%29