## College Algebra (6th Edition)

$$x_{1} = \frac{3 + \sqrt {15}}{3}$$ $$x_{2} = \frac{3 - \sqrt {15}}{3}$$
We have $$3x^2 - 6x - 2 = 0$$ As we see is a quadratic equation so we are going to use the quadratic formula in order to solve it. In this case $a = 3$, $b = -6$ and $c = -2$ $$x_{1,2} = \frac{-b ± \sqrt {b^2 -4ac}}{2a} = \frac{-(-6) ± \sqrt {(-6)^2 -4(3)(-2)}}{2(3)}$$ Doing each operations taking in count the sings law $$x_{1,2} = \frac{6 ± \sqrt {36 + 24}}{6} = \frac{6 ± \sqrt {60}}{6}$$ $60$ does not have exact square root but we can simplify it knowing that $60 = 4*15$ and square root of $4$ is $2$ so $\sqrt {60} = \sqrt {4*15} = 2\sqrt {15}$. Then $$x_{1,2} = \frac{6 ± 2\sqrt {15}}{6}$$ For the first value of $x$ we have $$x_{1} = \frac{6 + 2\sqrt {15}}{6} = \frac{3 + \sqrt {15}}{3}$$ And for the second value of $x$ we have $$x_{2} = \frac{6 - 2\sqrt {15}}{6} = \frac{3 - \sqrt {15}}{3}$$