What are the x intercepts of f(x)=x^2+7x+2 ?

For this case we have a function of the form:[tex]y = f (x)[/tex]Where:[tex]f (x) = x ^ 2 + 7x + 2[/tex]To find the intersections with the x-axis, we make [tex]y = 0[/tex], that is:[tex]x ^ 2 + 7x + 2 = 0[/tex]Where:[tex]a = 1\\b = 7\\c = 2[/tex]To find the solution we apply the following formula:[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]Substituting:[tex]x = \frac {-7 \pm \sqrt {7 ^ 2-4 (1) (2)}} {2 (1)}\\x = \frac {-7 \pm \sqrt {49-8}} {2}\\x = \frac {-7 \pm \sqrt {41}} {2}[/tex]We have two roots:[tex]x_ {1} = \frac {-7+ \sqrt {41}} {2}\\x_ {2} = \frac {-7- \sqrt {41}} {2}[/tex]Thus, the intersections with the "x" axis are:[tex](\frac {-7+ \sqrt {41}} {2}, 0)\\(\frac {-7- \sqrt {41}} {2}, 0)[/tex]Answer:[tex](\frac {-7+ \sqrt {41}} {2}, 0)\\(\frac {-7- \sqrt {41}} {2}, 0)[/tex]