SOLUTION: solve for x. x^2+18x+4=0

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Question 342463: solve for x. x^2+18x+4=0
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B18x%2B4=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B18x%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=%2818%29%5E2-4%2A1%2A4=308.

Discriminant d=308 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-18%2B-sqrt%28+308+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2818%29%2Bsqrt%28+308+%29%29%2F2%5C1+=+-0.225035612607877
x%5B2%5D+=+%28-%2818%29-sqrt%28+308+%29%29%2F2%5C1+=+-17.7749643873921

Quadratic expression 1x%5E2%2B18x%2B4 can be factored:
1x%5E2%2B18x%2B4+=+1%28x--0.225035612607877%29%2A%28x--17.7749643873921%29
Again, the answer is: -0.225035612607877, -17.7749643873921. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B18%2Ax%2B4+%29