Question 179370
{{{(2x^2-3x)+1=0}}}
:
{{{2(x^2-(3/2)x)+1=0}}} factored a two out of the x terms
:
{{{2(x^2-(3/2)x+(9/16))+1-(9/16)=0}}} took coefficient in front of x term and divided by 2.(3/2)/2= 3/4. Then we square that value {{{(3/4)^2}}}=9/16 which is the value you add to complete the square , but then you have to subtract it back out, outside the parenthesis to keep and equivilent equation
:
{{{2(x-(3/4))^2+(7/16)=0}}}