PLEAS HEL

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

-3/2 ±1/2 sqrt(5) = x

Step-by-step explanation:

f(x) = 2x^2 + 6x + 2

Set the function equal to 0

0 = 2x^2 + 6x + 2

Factor out 2

0 = 2(x^2 + 3x + 1)

Divide by 2

0 = (x^2 + 3x + 1)

Subtract 1 from each side

-1 = x^2 +3x

Complete the square by dividing the x coefficient by 2 and then squaring

(3/2)^2 = 9/4

-1 +9/4 = x^2 +3x+9/4

-4/4+9/4 = (x+3/2) ^2

5/4 = (x+3/2) ^2

Taking the square root of each side

± sqrt(5/4) = sqrt( (x+3/2) ^2)

±1/2 sqrt(5) =  (x+3/2)

Subtract 3/2 from each side

-3/2 ±1/2 sqrt(5) = x

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: