How do you solve 22x−x2=96?
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Answered by
0
Hey mate!
Here is yr answer.....
22x -x² = 96
-x² +22x -96 = 0
-x² +16x +6x -96 = 0
-x(x-16) +6(x-16) = 0
(x-16) (-x+6) = 0
(x-16) = 0 => x = 16
(-x+6) = 0 => x = 6
Hope it helps...
Here is yr answer.....
22x -x² = 96
-x² +22x -96 = 0
-x² +16x +6x -96 = 0
-x(x-16) +6(x-16) = 0
(x-16) (-x+6) = 0
(x-16) = 0 => x = 16
(-x+6) = 0 => x = 6
Hope it helps...
Answered by
0
22x-x^2=96
x^2-22x+96=0
x^2-16x-6x+96=0
x(x-16)-6(x-16)=0
(x-16)(x-6)=0
x=16 or x=6
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