## Algebra 1

$c=-10$
Given: $4(c+25)=c(c+4)$ $4c+100=c^2+4c$ $c^2+4c-4c-100=0$ $c^2-100=0$ $(c^2-10^2)=0$ As we know, $(a^2-b^2)=(a-b)(a+b)$ $(c-10)(c+10)=0$ Solutions are: $c=10,-10x=3,-1$ For $c=10$ the equation does not satisfied so the suitable answer is $c=-10$