How do you solve #22x-x^2=96#?

Question

3308 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-15T12:47:38

# x = 16 or 6#

#22x -x^2 =96#

#-x^2+22x-96=0#

#x^2 -22x +96=0#

#(x-16)(x-6)=0#

Hence, #x= 16 or 6#

Answer 2 · 2018-03-15T12:52:03

#x=6 and 16#

So first we want to get the #x^2# to be positive. To do this we have to multiply both sides by a #(-1)#.
#(-1)[22x-x^2=96]#

Giving us
#-22x+x^2=-96#

Now we can get all numbers to one side of the equation making it equal to #0#.
#x^2-22x+96=0#

Now we can factor. Since #-16*-6=96#
We can now write #x^2-22x+96#

As
#(x-16)(x-6)=0#

Setting both to #0# gives us
#x-16=0# and #x-6=0#

Which gives us
#x=16# and #x=6#