Question 992366
{{{3x^2-15x-7=0}}}... group first two terms together


{{{(3x^2-15x)-7=0}}}...factor out {{{3}}}


{{{3(x^2-5x)-7=0}}}.........complete the square of {{{(x^2-5x)}}}


to do it, recall that {{{(a-b)^2=a^2-2ab+b^2}}}
so, we need to add {{{b^2}}} inside parentheses then subtract it outside parentheses
we do not know {{{b}}}, but we know {{{a=1}}} and {{{2ab=5}}} and we can find {{{b}}}


{{{2ab=5}}}=>{{{2*1*b=5}}}=>{{{b=5/2}}} and {{{b^2=(5/2)^2}}}

now add it to {{{3(x^2-5x)-7=0}}}, and do not forget to multiply by {{{3}}} ones you subtract {{{(5/2)^2}}} because {{{3}}} is in front of parentheses


{{{3(x^2-5x+(5/2)^2)-3(5/2)^2-7=0}}}


{{{3(x-(5/2)^2)-3(25/4)-7=0}}}


{{{3(x-(5/2)^2)-75/4-7=0}}}


{{{3(x-(5/2)^2)-(75/4+7*4/4)=0}}}


{{{3(x-(5/2)^2)-(75/4+28/4)=0}}}


{{{3(x-5/2)^2-103/4=0}}}