-x+x^2= 56???
does this become
-x+x^2-56
what do you do after this
2007-03-11 09:18:24 · 6 answers · asked by :) 3 in Science & Mathematics ➔ Mathematics
-x + x^2 = 56
First, move the 56 to the left hand side.
-x + x^2 - 56 = 0
Now, rearrange this such that it is in descending power of x.
x^2 - x - 56 = 0
Now factor as normal.
(x - 8)(x + 7) = 0
Which means x = {8, -7}
2007-03-11 09:23:18 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
-x+x^2= 56 becomes -x x^2 -56 = 0
or x^2 - x -56 =0 which is a quadratic equation.
2007-03-11 16:21:56 · answer #2 · answered by physicist 4 · 0⤊ 0⤋
write the x^2 first
x^2-x-56=0
(x+7)(x-8)
x=-7, x=8
2007-03-11 16:20:59 · answer #3 · answered by leo 6 · 0⤊ 0⤋
Put it in order
x^2 - x - 56 = 0
(x -8)(x +7) = 0
x = 8
x = -7
2007-03-11 16:21:48 · answer #4 · answered by ecolink 7 · 0⤊ 0⤋
x^2 - x - 56 = 0
(x - 8)(x + 7) = 0
x = 8, -7
2007-03-11 16:21:27 · answer #5 · answered by Newbody 4 · 0⤊ 0⤋
It becomes (x-8)(x+7)=0
2007-03-11 16:46:06 · answer #6 · answered by Anonymous · 0⤊ 0⤋