Answer:
The answer is the option B
\(-4,-14\)
Step-by-step explanation:
we know that
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
\(x^{2}+18x+81=25\)
\(x^{2}+18x+56=0\)
so
\(a=1\\b=18\\c=56\)
substitute in the formula
\(x=\frac{-18(+/-)\sqrt{18^{2}-4(1)(56)}} {2(1)}\)
\(x=\frac{-18(+/-)\sqrt{100}} {2}\)
\(x=\frac{-18(+/-)10} {2}\)
\(x1=\frac{-18+10} {2}=-4\)
\(x2=\frac{-18-10} {2}=-14\)
