Math, asked by naman7514, 11 months ago

How do you solve 22x−x2=96?

Answers

Answered by Anonymous
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...


Answered by anju7699
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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