Question 1047286


{{{2x^2-5x=7}}}

{{{2x^2-5x-7=0}}} ....to complete square, write {{{-5x}}} as {{{2x-7x}}}

{{{2x^2+2x-7x-7=0}}}....group

{{{(2x^2+2x)-(7x+7)=0}}}...factor out {{{2x}}} and {{{7}}}

{{{2x(x+1)-7(x+1)=0}}}

{{{(2x-7)(x+1)=0}}}

so, the solution to this will be:

if {{{(2x-7)=0}}}->{{{2x=7}}}->{{{x=7/2}}}->{{{x=3.5}}}
if {{{(x+1)=0}}}->{{{x=-1}}}


{{{ graph( 600, 600, -10, 10, -10, 10, 2x^2-5x-7) }}}