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

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Answer 1 · 2017-02-08T16:00:28

The solutions are $S = {\frac{2}{3} + \frac{1}{3} \sqrt{2} i, \frac{2}{3} - \frac{1}{3} \sqrt{2} i}$ $S={2/3+1/3sqrt2i,2/3-1/3sqrt2i}$

We compare this equation to

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$3 x^{2} + 2 = 4 x$ $3x^2+2=4x$

$3 x^{2} - 4 x + 2 = 0$ $3x^2-4x+2=0$

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 $CC$

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

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

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

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

How do you solve 3x^2+2=4x3x2+2=4x3x^2+2=4x using the quadratic formula?