Answer:
The zeros are
[tex]x1=\frac{-5+\sqrt{5}} {2}[/tex] and [tex]x2=\frac{-5-\sqrt{5}} {2}[/tex]
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]f(x)=x^{2} +5x+5[/tex]
To find the zeros equate the function to 0
[tex]x^{2} +5x+5=0[/tex]
so
[tex]a=1\\b=5\\c=5[/tex]
substitute in the formula
[tex]x=\frac{-5(+/-)\sqrt{5^{2}-4(1)(5)}} {2(1)}[/tex]
[tex]x=\frac{-5(+/-)\sqrt{5}} {2}[/tex]
[tex]x1=\frac{-5+\sqrt{5}} {2}[/tex]
[tex]x2=\frac{-5-\sqrt{5}} {2}[/tex]
