SOLUTION: The problem is 5x^2=-12x-5 Don't I set this to "0"? I moved the -12x-5 over to the left (the other side of the =) to read: 5x^2+12x+5=0 Is this the final answer? I get

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Question 531111: The problem is 5x^2=-12x-5
Don't I set this to "0"?
I moved the -12x-5 over to the left (the other side of the =) to read:
5x^2+12x+5=0
Is this the final answer? I get confused on if I need to do anymore or if this should be the end of the problem.
Your help is greatly appreciated.
Cheers!
Angela
Answer by oberobic(2304) (Show Source):
You can put this solution on YOUR website!
Hint: You cannot "move" anything when dealing with equations. You must do the same thing to both sides of the equation. The notion of "moving" obscures that you are doing something to both sides.
.
5x^2 = -12x + 5
.
Add 12x to both sides.
.
5x^2 +12x = (-12x+12x) + 5
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Subtract 5 from both sides.
.
5x^2 +12x -5 = 0
.

Solved by pluggable solver: SOLVE quadratic equation with variable

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=244 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.362049935181331, -2.76204993518133. Here's your graph: