If $x=18$ is one of the solutions of the equation $x^{2}+12 x+c=0$, the other solution of this equation is
(A) $x=216$
(B) $x=-6$
(C) $x=-30$
(D) $x=30$
(E) $x=-540$
Solution 1

Since $x=18$ is a solution to the equation $x^{2}+12 x+c=0$, then $x=18$ satisfies this equation. Thus, $18^{2}+12(18)+c=0$ and so $324+216+c=0$ or $c=-540$.

Therefore, the original equation becomes $x^{2}+12 x-540=0$ or $(x-18)(x+30)=0$.

Therefore, the other solution is $x=-30$.

## Solution 2

We use the fact that the sum of the roots of an equation of the form $x^{2}+b x+c=0$ is $-b$. If the roots of the equation $x^{2}+12 x+c=0$ are 18 and $r$, then $18+r=-12$ or $r=-30$. Therefore, the other solution is $x=-30$.