Question 115423
{{{2x^2+10x+11=0}}} Start with the given equation



{{{2x^2+10x=-11}}} Subtract 11 from both sides
{{{2(x^2+5x)=-11}}} Factor out the leading coefficient 2.  This step is important since we want the {{{x^2}}} coefficient to be equal to 1.




Take half of the x coefficient 5 to get {{{5/2}}}

Now square {{{5/2}}} to get {{{25/4}}}




{{{2(x^2+5x+25/4)=-11+(25/4)(2)}}} Add this result {{{25/4}}} to the expression {{{x^2+5x}}}  inside the parenthesis. Now the expression {{{x^2+5x+25/4}}}  is a perfect square trinomial. Now add the result {{{(25/4)(2)}}} (remember we factored out a 2) to the right side.




{{{2(x+5/2)^2=-11+(25/4)(2)}}} Factor {{{x^2+5x+25/4}}} into {{{(x+5/2)^2}}} 



{{{2(x+5/2)^2=-11+25/2}}} Multiply  {{{25/4}}} and 2 to get  {{{25/2}}}




{{{2(x+5/2)^2=3/2}}} Combine like terms on the right side


{{{(x+5/2)^2=3/4}}} Divide both sides by 2



{{{x+5/2=0+-sqrt(3/4)}}} Take the square root of both sides



{{{x+5/2=0+-sqrt(3)/sqrt(4)}}} Break up the square root



{{{x+5/2=0+-sqrt(3)/2}}} Take the square root of 4 to get 2



{{{x=-5/2+-sqrt(3)/2}}} Subtract {{{5/2}}} from both sides to isolate x.



{{{x=(-5+-sqrt(3))/2}}} Combine the fractions




So our answer is 

{{{x=(-5+sqrt(3))/2}}}  or {{{x=(-5-sqrt(3))/2}}}