SOLUTION: I need to find out how to work this problem y2-2y-4=0

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Question 532147: I need to find out how to work this problem y2-2y-4=0
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
I assume you mean
+y%5E2+-2y+-4+=+0+
.
Use the quadratic equation
.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 1y%5E2%2B-2y%2B-4+=+0) has the following solutons:

y%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=%28-2%29%5E2-4%2A1%2A-4=20.

Discriminant d=20 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+20+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+20+%29%29%2F2%5C1+=+3.23606797749979
y%5B2%5D+=+%28-%28-2%29-sqrt%28+20+%29%29%2F2%5C1+=+-1.23606797749979

Quadratic expression 1y%5E2%2B-2y%2B-4 can be factored:
1y%5E2%2B-2y%2B-4+=+1%28y-3.23606797749979%29%2A%28y--1.23606797749979%29
Again, the answer is: 3.23606797749979, -1.23606797749979. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-4+%29