How do you solve $22 x - x^{2} = 96$ $22x-x^2=96$ ?

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How do you solve $22 x - x^{2} = 96$ $22x-x^2=96$ ?

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Answer 1 · 2018-03-15T12:47:38

$x = 16 or 6$ $x = 16 or 6$

Explanation:

$22 x - x^{2} = 96$ $22x -x^2 =96$

$- x^{2} + 22 x - 96 = 0$ $-x^2+22x-96=0$

$x^{2} - 22 x + 96 = 0$ $x^2 -22x +96=0$

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

Hence, $x = 16 or 6$ $x= 16 or 6$

Answer 2 · 2018-03-15T12:52:03

$x = 6 and 16$ $x=6 and 16$

Explanation:

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

Giving us
$- 22 x + x^{2} = - 96$ $-22x+x^2=-96$

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

Now we can factor. Since $- 16 \cdot - 6 = 96$ $-16*-6=96$
We can now write $x^{2} - 22 x + 96$ $x^2-22x+96$

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

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

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

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How do you solve $22 x - x^{2} = 96$ $22x-x^2=96$ ?

2 Answers

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How do you solve 22x−x2=9622x-x^2=96?

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How do you solve $22 x - x^{2} = 96$ $22x-x^2=96$ ?