Answer:
The solution set is [tex]x=-9[/tex]
Step-by-step explanation:
we have
[tex](x+9)(x+9)=0[/tex]
we know that
[tex](x+9)(x+9)=x^{2}+18x+81[/tex]
so
[tex]x^{2}+18x+81=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]x^{2}+18x+81=0[/tex]
so
[tex]a=1\\b=18\\c=81[/tex]
substitute in the formula
[tex]x=\frac{-18(+/-)\sqrt{18^{2}-4(1)(81)}} {2(1)}[/tex]
[tex]x=\frac{-18(+/-)\sqrt{0}} {2}[/tex]
The radicand is equal to zero, therefore, the solution has only one real solution
[tex]x=\frac{-18} {2}=-9[/tex]
