SOLUTION: Use the quadratic formula to solve the equation. (x+4)(x-2)=5

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Question 961950: Use the quadratic formula to solve the equation.
(x+4)(x-2)=5
Answer by amarjeeth123(569) About Me  (Show Source):
You can put this solution on YOUR website!
(x+4)(x-2)=5
x^2+4x-2x-8=5
x^2+2x-13=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-13+=+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=%282%29%5E2-4%2A1%2A-13=56.

Discriminant d=56 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+56+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+56+%29%29%2F2%5C1+=+2.74165738677394
x%5B2%5D+=+%28-%282%29-sqrt%28+56+%29%29%2F2%5C1+=+-4.74165738677394

Quadratic expression 1x%5E2%2B2x%2B-13 can be factored:
1x%5E2%2B2x%2B-13+=+1%28x-2.74165738677394%29%2A%28x--4.74165738677394%29
Again, the answer is: 2.74165738677394, -4.74165738677394. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-13+%29