How do you solve #3x^2+2=4x# using the quadratic formula?

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Answer 1 · 2017-02-08T16:00:28

The solutions are #S={2/3+1/3sqrt2i,2/3-1/3sqrt2i}#

We compare this equation to

#ax^2+bx+c=0#

#3x^2+2=4x#

#3x^2-4x+2=0#

We calculate the discriminant

#Delta=b^2-4ac#

#Delta=(-4)^2-4(3)(2)=16-24=-8#

As, #Delta<0#, the solutions are not in #RR# but in #CC#

#x=(-b+-sqrtDelta)/(2a)#

#x=(4+-sqrt(-8))/6#

#x_1=(4+2sqrt2i)/6=2/3+1/3sqrt2i#

#x_2=(4-2sqrt2i)/6=2/3-1/3sqrt2i#