Question 169644
{{{3x^2=-12x-1}}} Start with the given equation



{{{3x^2+12x+1=0}}} Add 12x to both sides. Add 1 to both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{b=12}}}, 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 = (-(12) +- sqrt( (12)^2-4(3)(1) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=12}}}, and {{{c=1}}}



{{{x = (-12 +- sqrt( 144-4(3)(1) ))/(2(3))}}} Square {{{12}}} to get {{{144}}}. 



{{{x = (-12 +- sqrt( 144-12 ))/(2(3))}}} Multiply {{{4(3)(1)}}} to get {{{12}}}



{{{x = (-12 +- sqrt( 132 ))/(2(3))}}} Subtract {{{12}}} from {{{144}}} to get {{{132}}}



{{{x = (-12 +- sqrt( 132 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-12 +- 2*sqrt(33))/(6)}}} 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 = (-12+2*sqrt(33))/(6)}}} or {{{x = (-12-2*sqrt(33))/(6)}}} Break up the expression.  



So the answers are {{{x = (-12+2*sqrt(33))/(6)}}} or {{{x = (-12-2*sqrt(33))/(6)}}} 



which approximate to {{{x=-0.085}}} or {{{x=-3.915}}}