SOLUTION: How can I solve (x+3)(x+4)=0 and 2x(x-3)=15???? In standard form??

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Question 1083943: How can I solve (x+3)(x+4)=0 and 2x(x-3)=15????
In standard form??
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How can I solve (x+3)(x+4)=0 and 2x(x-3)=15????
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(x+3)(x+4)=0
If the product = 0, either
x+3 = 0 --> x = -3
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or x+4 = 0 --> x = -4
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2x(x-3)=15
2x^2 - 6x - 15 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-6x%2B-15+=+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=%28-6%29%5E2-4%2A2%2A-15=156.

Discriminant d=156 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+156+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+156+%29%29%2F2%5C2+=+4.6224989991992
x%5B2%5D+=+%28-%28-6%29-sqrt%28+156+%29%29%2F2%5C2+=+-1.6224989991992

Quadratic expression 2x%5E2%2B-6x%2B-15 can be factored:
2x%5E2%2B-6x%2B-15+=+%28x-4.6224989991992%29%2A%28x--1.6224989991992%29
Again, the answer is: 4.6224989991992, -1.6224989991992. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-6%2Ax%2B-15+%29