Question 184810


{{{18x^2+9x+1=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=18}}}, {{{b=9}}}, and {{{c=1}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(9) +- sqrt( (9)^2-4(18)(1) ))/(2(18))}}} Plug in  {{{a=18}}}, {{{b=9}}}, and {{{c=1}}}



{{{x = (-9 +- sqrt( 81-4(18)(1) ))/(2(18))}}} Square {{{9}}} to get {{{81}}}. 



{{{x = (-9 +- sqrt( 81-72 ))/(2(18))}}} Multiply {{{4(18)(1)}}} to get {{{72}}}



{{{x = (-9 +- sqrt( 9 ))/(2(18))}}} Subtract {{{72}}} from {{{81}}} to get {{{9}}}



{{{x = (-9 +- sqrt( 9 ))/(36)}}} Multiply {{{2}}} and {{{18}}} to get {{{36}}}. 



{{{x = (-9 +- 3)/(36)}}} Take the square root of {{{9}}} to get {{{3}}}. 



{{{x = (-9 + 3)/(36)}}} or {{{x = (-9 - 3)/(36)}}} Break up the expression. 



{{{x = (-6)/(36)}}} or {{{x =  (-12)/(36)}}} Combine like terms. 



{{{x = -1/6}}} or {{{x = -1/3}}} Simplify. 



So the answers are {{{x = -1/6}}} or {{{x = -1/3}}} 

  

Note: the order of the solutions does not matter