Answer:
The zero's of the function are \(x=-6\) and \(x=-7\)
Step-by-step explanation:
we have
\(h(w)=w^{2} +13w+42\)
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
\(w^{2} +13w+42=0\)
The formula to solve a quadratic equation of the form \(ax^{2} +bx+c=0\) is equal to
\(x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}\)
in this problem we have
\(w^{2} +13w+42=0\)
so
\(a=1\\b=13\\c=42\)
substitute in the formula
\(x=\frac{-13(+/-)\sqrt{13^{2}-4(1)(42)}} {2(1)}\)
\(x=\frac{-13(+/-)\sqrt{169-168}} {2}\)
\(x=\frac{-13(+/-)1} {2}\)
\(x=\frac{-13(+)1} {2}=-6\)
\(x=\frac{-13(-)1} {2}=-7\)
