SOLUTION: Please help me solve the following equation: (2x-5)(x+8)=0 This is as far as I have gotten: 2x^2+16x-5x-40=0 2x^2+11x-40=0 2x^2+11x=40

Question 162831: Please help me solve the following equation:
(2x-5)(x+8)=0
This is as far as I have gotten:
2x^2+16x-5x-40=0
2x^2+11x-40=0
2x^2+11x=40

Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
You did not need to expand it. This, is enough:
(2x-5)(x+8)=0
.
Notice that if EITHER then contents of the left parenthesis or the right parenthesis, is zero, then the ENTIRE left side is zero.
.
Therefore, the solutions are found by setting the two term on the left to zero.
Solution 1:
(2x-5)=0
2x = 5
x = 5/2
.
Solution 2:
(x+8)=0
x = -8
.
Therefore,
x = {-8, 5/2}
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