Question

PLEASE PLEASEEEEEEE HELPPPPPPPP

what are the zeros in this equation?
f(x) = 2x^2 + 6x^2 + 2

Answer 1

So you definately wrote something wrong but I will assume the 6x^2 was meant to be just 6x.

2x^2 + 6x + 2 = 0

x^2 + 3x + 1 = 0

I can't seem to find a better way than to simply use the quadratic formula:

(-3 + sqrt(9 - 4*1*1))/2
`\approx` -0.382

(-3 - sqrt(9 - 4*1*1))/2
`\approx` -2.62

So the zeroes for this function are at

(-0.382, 0) and (-2.62, 0)

Or exactly:

((-3 + sqrt(9 - 4*1*1))/2, 0) and ((-3 - sqrt(9 - 4*1*1))/2, 0)