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