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

Question 1083943: How can I solve (x+3)(x+4)=0 and 2x(x-3)=15????
In standard form??

Answer by Alan3354(69443) (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 (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=156 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 4.6224989991992, -1.6224989991992. Here's your graph:

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