Question 211966
F(x) = 3x^2-2x+8

Now first we set F(x) = 4 and solve
3x^2-2x+8 = 4
3x^2-2x+4 = 0

Now try as we might this equation won't factor so we are going to have to solve using the quadratic formula.  
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

where a = 3, b = -2, and c = 4 so pluging in and simplifying we get:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (2 +- sqrt( 2^2-4*3*4 ))/(2*3) }}}
{{{x = (2 +- sqrt( 4-48 ))/(6) }}}
{{{x = (2 +- sqrt( -44 ))/(6) }}}

Now if you don't know about imaginary numbers this is where we stop because you can't take the square root of a negative number with out getting an imaginary number.  Otherwise we continue.  

{{{x = (2 +- 2sqrt( 11 )i)/(6) }}}
{{{x = (1 +- sqrt( 11 )i)/(3) }}}

and thats your final answer.

You do the same thing with the next polynomial, just simplify and then plug and chug through the quadratic formula.  Again your answer will have imaginary numbers in it.