Question

what is the root of[9x-8=x^{2}

Answer 1

positive 8 and 1

Answer 2

You subtract the 9x and add the 8 to the other side withx^2

Answer 3

then, you factor that to get (x-1) and (x-8)

Answer 4

after that, set them to equal zero like so: x-1= 0
and x-8= 0

Answer 5

what is the roots for these 3problems
x^{2}-7x-4=10 x^{2}-20x+99=0 x^{2}=x-72=0

Answer 6

for the first one add 10 to the opposite side.

Answer 7

you notice that these do not have any possible multiples to equal to 14

Answer 8

so what you do is, take the equation : b+ or - the square root of b^2 -4(a)(c)

Answer 9

that will give you -7+or- the square root of -7^2-4(1)(14)

Answer 10

that will be -7+or- the square root of 7

Answer 11

the final answer for that one is -7+or- the square root of 7 over 2.

I forgot to tell you that the equation I posted ealier:b+ or - the square root of b^2 -4(a)(c) - is divided by two. You use this equation when the mulitples of the third term in a problem cannot give you the middle term when added or subtracted