How to factor this????

Question

x^2-7x=-10 ??? Isn't it prime???

Answer 1

x^2 - 7x = -10 can be written as

x^ - 7x + 10 = 0 and this can be factored as

x - 5 and x - 2.

Check that - 5 - 2 = - 7 and - 5 x - 2 = + 10

So, (x - 5) (x - 2) = 0

and that means x = 5 and x = 2 are the two values of x.

Answer 2

I will give you something to remember how to factor whether quadric question ...

a*x^2+b*x+c=0

now what is a

a=1 as x^2=a*x^2
b=-7 as -7*x=b*x
c=10 (x^2-7x=-10 ---> x^2-7x+10=-10+10 -->x^2-7x+10=0)

now
x=(-b-(b^2-4*a*c))/2 or (-b+(b^2-4*a*c))/2

I guess you mean with prime that is solvable with real roots not imaginary ones, so:

check out b^2-4*a*c<0 not solvable, >0 solvable

(-7)^2-4*1*10=49-40=9 >0 then it is solvable

now
(-b+SQRT(9))/(2*1)=(7+3)/2 =5
the other root
(-b-SQRT(9))/(2*1)=(7-3)/2 =2

That's all,

I note you still sad, I hope you will be someday happy
salud

Answer 3

x^2 - 7x + 10 = 0

(x-5)(x-2) = 0

x = 5 or 2