SOLUTION: how do you solve (3-y)(y+4)=3y-5

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Question 765432: how do you solve (3-y)(y+4)=3y-5
Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
(3-y)(y+4)=3y-5
3y+12-y^2-4y=3y-5
y^2+4y-17=

then...

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 1y%5E2%2B4y%2B-17+=+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=%284%29%5E2-4%2A1%2A-17=84.

Discriminant d=84 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+84+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%284%29%2Bsqrt%28+84+%29%29%2F2%5C1+=+2.58257569495584
y%5B2%5D+=+%28-%284%29-sqrt%28+84+%29%29%2F2%5C1+=+-6.58257569495584

Quadratic expression 1y%5E2%2B4y%2B-17 can be factored:
1y%5E2%2B4y%2B-17+=+1%28y-2.58257569495584%29%2A%28y--6.58257569495584%29
Again, the answer is: 2.58257569495584, -6.58257569495584. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-17+%29