## College Algebra (6th Edition)

$$x_{1}= -3 + 2\sqrt {6}$$ $$x_{2}= -3 - 2\sqrt {6}$$
We have $$(x+3)^2 = 24$$ Squaring the binomial expression $$(x)^2 + 2(x)(3) + (3)^2 = 24$$ $$x^2 + 6x + 9 = 24$$ As we see we got a quadratic term, a lineal term and constants so we are going to move everything to the right side of the equation and equalize it to $0$ $$x^2 + 6x + 9 -24 = 0$$ $$x^2 + 6x + -15 = 0$$ We got a quadratic equation so we are going to solve it with quadratic formula. In this case $a = 1$, $b = 6$ and $c = -15$ $$x_{1,2} = \frac{-b ± \sqrt {b^2 - 4ac}}{2a} = \frac{-6 ± \sqrt {(6)^2 - 4(1)(-15)}}{2(1)}$$ Doing the operations taking in count the signs law $$x_{1,2} = \frac{-6 ± \sqrt {36 +60}}{2} = \frac{-6 ± \sqrt {96}}{2}$$ $96$ does not has exact square root but we can say that $96 = 16*6$ and $16$ has exact square root so $\sqrt {96} = \sqrt {16*6} = 4\sqrt {6}$. Then we get $$x_{1,2}= \frac{-6 ± 4\sqrt {6}}{2}$$ For the first value of $x$ we have $$x_{1} = \frac{-6 + 4\sqrt {6}}{2} = -3 + 2\sqrt {6}$$ And for the second value of $x$ we have $$x_{2} = \frac{-6 - 4\sqrt {6}}{2} = -3 - 2\sqrt {6}$$