Question 411320
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Hi
finding the roots using the quadratic formula
3y^2– 2y + 12 = 0
 {{{x = (-3 +- sqrt( -140 ))/(6) }}}
 {{{x = (-3 +- i*sqrt( 4*35 ))/(6) }}}
 {{{x = (-3 +- 2i*sqrt(35))/(6) }}}
    x = -.5 + 2isqrt(35)/6
    x = -.5 - 2isqrt(35)/6
Sum of the roots = -1
Product: .25 + 4*35/36 
Note i^2 = -1