I NEED HELP ASAP THISIS WORTH 3 GRADES

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0

given the equation is 8+9z=-6, the first step would be to subtract the 8 from both sides:

8+9z-8=-6-8

since 8 minus 8 = 0, this would cancel out the 8. and since -6 minus 8 equals -14, the equation would look like this

9z = -14