Question 943794

if you have {{{2x^2-7x }}}, to find solutions (you have second degree function, so you will have two solutions) first make it equal to zero


{{{2x^2-7x=0}}} ...factor out {{{x}}}


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

solutions:

{{{x(2x-7)=0}}} will be true if =>  {{{x=0}}} or {{{(2x-7)=0}}}

so,the first solution is {{{x=0}}}

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


so, the second solution is {{{x=3.5}}}