Question 170019
Are you sure that the equation is not {{{2x^2+11x+5=0}}}??



If so, then...



{{{2x^2+11x+5=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=11}}}, and {{{c=5}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(11) +- sqrt( (11)^2-4(2)(5) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=11}}}, and {{{c=5}}}



{{{x = (-11 +- sqrt( 121-4(2)(5) ))/(2(2))}}} Square {{{11}}} to get {{{121}}}. 



{{{x = (-11 +- sqrt( 121-40 ))/(2(2))}}} Multiply {{{4(2)(5)}}} to get {{{40}}}



{{{x = (-11 +- sqrt( 81 ))/(2(2))}}} Subtract {{{40}}} from {{{121}}} to get {{{81}}}



{{{x = (-11 +- sqrt( 81 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-11 +- 9)/(4)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{x = (-11 + 9)/(4)}}} or {{{x = (-11 - 9)/(4)}}} Break up the expression. 



{{{x = (-2)/(4)}}} or {{{x =  (-20)/(4)}}} Combine like terms. 



{{{x = -1/2}}} or {{{x = -5}}} Simplify. 



So the answers are {{{x = -1/2}}} or {{{x = -5}}}