Question 147093
{{{2 x^2-4 x-7}}} Start with the given expression



{{{2(x^2-2x-7/2)}}} Factor out the leading coefficient {{{2}}}



Take half of the x coefficient {{{-2}}} to get {{{-1}}} (ie {{{(1/2)(-2)=-1}}}).


Now square {{{-1}}} to get {{{1}}} (ie {{{(-1)^2=(-1)(-1)=1}}})





{{{2(x^2-2x+1-1-7/2)}}} Now add and subtract this value inside the parenthesis. Notice how {{{1-1=0}}}. Since we're adding 0, we're not changing the equation.




{{{2((x-1)^2-1-7/2)}}} Now factor {{{x^2-2x+1}}} to get {{{(x-1)^2}}}



{{{2((x-1)^2-9/2)}}} Combine like terms



{{{2(x-1)^2+2(-9/2)}}} Distribute



{{{2(x-1)^2-9}}} Multiply




So after completing the square, {{{2x^2-4x-7}}} becomes {{{2(x-1)^2-9}}}.



In other words, {{{2x^2-4x-7=2(x-1)^2-9}}}


So {{{2x^2-4x-7=0}}} is equivalent to {{{2(x-1)^2-9=0}}}



{{{2(x-1)^2-9=0}}} Start with the given equation



{{{2(x-1)^2=9}}} Add 9 to both sides



{{{(x-1)^2=9/2}}} Divide both sides by 2



{{{x-1=0+-sqrt(9/2)}}} Take the square root of both sides



{{{x=1+-sqrt(9/2)}}} Add 1 to both sides



{{{x=1+sqrt(9/2)}}} or {{{x=1-sqrt(9/2)}}} Break up the plus/minus



{{{x=1+3/sqrt(2)}}} or {{{x=1-3/sqrt(2)}}} Simplify



{{{x=1+3sqrt(2)/2}}} or {{{x=1-3sqrt(2)/2}}} Rationalize the denominator



So our answers are {{{x=1+3sqrt(2)/2}}} or {{{x=1-3sqrt(2)/2}}}