Answer:
[tex]x=1,-5[/tex]
Step-by-step explanation:
[tex]We\ are\ given\ that,\\4x+5=x(x+8)\\Hence\ by\ expanding\ the\ terms,\\4x+5=x^2+8x\\Hence\ by\ moving\ 4x\ and\ 5 to\ the\ RHS\ ,\\0=x^2+8x-4x-5\\Hence\ by\ simplifying\ the\ RHS,\\0=x^2+4x-5\\Now,\\We\ take\ a\ closer\ look\ at\ the\ RHS,\\x^2+4x-5\ can\ also\ be\ written\ as:\\=x^2+5x-x-5\\Now, By\ factorizing,\\x(x+5)-1(x+5)\\=(x-1)(x+5)\\Hence,\\(x-1)(x+5)=0\\Now,\\To\ nullify\ the\ LHS\ to\ 0,\ either\ the\ term\ (x-1)\ or/and\ (x+5)\ should\ equate\ to\ 0.\\Hence,\\x-1=0\\[/tex][tex]x=1\ is\ our\ first\ solution\\(x+5)=0\\x=-5\ is\ our\ second\ solution\\Hence,\\x=1,-5[/tex]
