Answer:
The zero's of the function are [tex]x=-6[/tex] and [tex]x=-7[/tex]
Step-by-step explanation:
we have
[tex]h(w)=w^{2} +13w+42[/tex]
we know that
The zero's of the function are the values of x when the value of the function is equal to zero
so
equate the function to zero
[tex]w^{2} +13w+42=0[/tex]
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]w^{2} +13w+42=0[/tex]
so
[tex]a=1\\b=13\\c=42[/tex]
substitute in the formula
[tex]x=\frac{-13(+/-)\sqrt{13^{2}-4(1)(42)}} {2(1)}[/tex]
[tex]x=\frac{-13(+/-)\sqrt{169-168}} {2}[/tex]
[tex]x=\frac{-13(+/-)1} {2}[/tex]
[tex]x=\frac{-13(+)1} {2}=-6[/tex]
[tex]x=\frac{-13(-)1} {2}=-7[/tex]
