Let k be the constant of correction
Write as #y=x^2+10x+5+k# At this stage k=0
#y=(x^(color(red)(2))+10x)+5+k#
Take the square from #x^(color(red)(2)# and move it outside the bracket
#y=(x+10x)^(color(red)(2))+5+k#
remove the #x# from #10x#
#y=(x+10)^2+5+k#
halve the 10
#y=(x+ color(magenta)(5))^2+5+k#
;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider #y=ax^2+bx+c# written in the form #y=a(x+color(magenta)(b/(2a)))^2+c+k#
From within the bracket we have #k= (-1)xx a xx(color(magenta)(b/(2a)))^2#
The #axx# is from the #a# outside the bracket but in your case #a=1#
so #k=(-1)xx(color(magenta)(+5))^2 = -25#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)(y=(x+5)^2+5+k)" "color(green)(->" "y=(x+5)^2+5-25#
#y=(x+5)^2-20#
'~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("solving for "x" when "y=0)#
#=>0=(x+5)^2-20#
#sqrt(20)=sqrt((x+5)^2)#
#=>x+5=+-2sqrt(5)#
#=>x=-5+-2sqrt(5)#
#color(blue)(x~~ -9.472" and "-0.528" to 3 decimal places")#
