SOLUTION: It's always been a pet hate of mine, but here's my question: How do you solve an equation when you have n^2 + n So for example 210 = n^2 + n. Find n

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Question 805402: It's always been a pet hate of mine, but here's my question:
How do you solve an equation when you have n^2 + n
So for example 210 = n^2 + n. Find n
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
210 = n^2 + n
0 = n^2 + n-210
n^2 + n-210=0
If you are lucky, you can factor or complete the square
(n+15)*(n-14) =0
n=-15 and n=14 are solutions
The method that always works is the quadratic equation. see below
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-210+=+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-210=841.

Discriminant d=841 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+841+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+841+%29%29%2F2%5C1+=+14
x%5B2%5D+=+%28-%281%29-sqrt%28+841+%29%29%2F2%5C1+=+-15

Quadratic expression 1x%5E2%2B1x%2B-210 can be factored:
1x%5E2%2B1x%2B-210+=+%28x-14%29%2A%28x--15%29
Again, the answer is: 14, -15. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-210+%29