Question 178398


{{{x^2+12x+11=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=12}}}, and {{{c=11}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(12) +- sqrt( (12)^2-4(1)(11) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=12}}}, and {{{c=11}}}



{{{x = (-12 +- sqrt( 144-4(1)(11) ))/(2(1))}}} Square {{{12}}} to get {{{144}}}. 



{{{x = (-12 +- sqrt( 144-44 ))/(2(1))}}} Multiply {{{4(1)(11)}}} to get {{{44}}}



{{{x = (-12 +- sqrt( 100 ))/(2(1))}}} Subtract {{{44}}} from {{{144}}} to get {{{100}}}



{{{x = (-12 +- sqrt( 100 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-12 +- 10)/(2)}}} Take the square root of {{{100}}} to get {{{10}}}. 



{{{x = (-12 + 10)/(2)}}} or {{{x = (-12 - 10)/(2)}}} Break up the expression. 



{{{x = (-2)/(2)}}} or {{{x =  (-22)/(2)}}} Combine like terms. 



{{{x = -1}}} or {{{x = -11}}} Simplify. 



So the answers are {{{x = -1}}} or {{{x = -11}}}