Question 171090

{{{2=-9x^2-x}}} Start with the given equation.



{{{2+9x^2+1x=0}}} Get all terms to the left side.



{{{9x^2+x+2=0}}} Rearrange the terms.



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



Let's use the quadratic formula to solve for x



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



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



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



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



{{{x = (-1 +- sqrt( -71 ))/(2(9))}}} Subtract {{{72}}} from {{{1}}} to get {{{-71}}}



{{{x = (-1 +- sqrt( -71 ))/(18)}}} Multiply {{{2}}} and {{{9}}} to get {{{18}}}. 



{{{x = (-1 +- i*sqrt(71))/(18)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-1+i*sqrt(71))/(18)}}} or {{{x = (-1-i*sqrt(71))/(18)}}} Break up the expression.  



So the answers are {{{x = (-1+i*sqrt(71))/(18)}}} or {{{x = (-1-i*sqrt(71))/(18)}}} 



which approximate to {{{x=-0.556+0.468*i}}} or {{{x=-0.5556-0.468*i}}}