How do you solve $3x² = 2x +4$ ?

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Answer 1 · 2017-01-20T14:59:57

First, rewrite it as $3x^2-2x-4=0$ , then use the quadratic formula and you find the roots to be $x=(1+-sqrt13)/3$

Once you have it in the form $3x^2-2x-4=0$ , the quadratic formula gives a reliable means of finding the roots:

$x=(2+- sqrt((-2)^2-4(3)(-4)))/(2(3))=(2+-sqrt52)/6=(2+-2sqrt13)/6$

Now, just divide each term by 2:

$x=(1+-sqrt13)/3$

How do you solve 3x² = 2x +43x² = 2x +4?