Answer
$$x_{1} = -3 + \sqrt 7 $$
$$x_{2} = -3 - \sqrt 7 $$
Work Step by Step
We have $$y = x^2 +6x + 2$$
To find the x-intercepts of the graphic we need to make $y=0$ and solve for $x$ so $$x^2+6x+2 = 0$$
As we see we have a quadratic equation so we are going to solve it with the quadratic formula. In this case $a = 1$, $b = 6$ and $c = 2$ $$x_{1,2} = \frac{-b ± \sqrt {b^2 - 4ac}}{2a} = \frac{-(6) ± \sqrt {(6)^2 - 4(1)(2)}}{2(1)}$$
Doing all the operations taking in count the signs law $$x_{1,2} = \frac{-6 ± \sqrt {36 - 8}}{2} =\frac{-6 ± \sqrt {28}}{2}$$
$28$ does not has exact square root but we can say that $28 = 4*7$ and 4 has square root, so $\sqrt {28} = \sqrt {4*7} = 2\sqrt 7$
Then we get $$x_{1,2} = \frac{-6 ± 2\sqrt {7}}{2}$$
Our first x-intercept is $$x_{1} = \frac{-6 + 2\sqrt {7}}{2} = -3 + \sqrt 7 $$
And our second x-intercept is $$x_{2} = \frac{-6 - 2\sqrt {7}}{2} = -3 - \sqrt 7 $$
