Question 147093
{{{2x^2-4x-7=0 }}}


Step 1:  Add the constant term to both sides


{{{2x^2-4x=7}}}


Step 2:  Divide both sides by the lead coefficient


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


Step 3:  Divide the coefficient on the 1st degree term by 2 ({{{(-2)/2}}}), square the result {{{(-1)^2=1}}}, then add it to both sides


{{{x^2-2x+1=7/2+1}}}


{{{x^2-2x+1=9/2}}}


Step 4:  Factor the perfect square on the left


{{{(x-1)^2=9/2}}}


Step 5:  Take the square root of both sides


{{{x-1=sqrt(9/2)}}} or {{{x-1=-sqrt(9/2)}}}


Step 6:  Add the constant term on the left to both sides


{{{x=1+-sqrt(9/2)}}}


Step 7:  Simplify


(Left as an exercise for the student.  Remember to rationalize your denominator)