Question 883291
I'm assuming you want to solve {{{3x^2+x+4=0}}}





{{{3x^2+x+4=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2+x+4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=1}}}, and {{{C=4}}}



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(3)(4) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=1}}}, and {{{C=4}}}



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



{{{x = (-1 +- sqrt( 1-48 ))/(2(3))}}} Multiply {{{4(3)(4)}}} to get {{{48}}}



{{{x = (-1 +- sqrt( -47 ))/(2(3))}}} Subtract {{{48}}} from {{{1}}} to get {{{-47}}}



{{{x = (-1 +- sqrt( -47 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-1 +- i*sqrt(47))/(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 = (-1+i*sqrt(47))/(6)}}} or {{{x = (-1-i*sqrt(47))/(6)}}} Break up the expression.  



So the solutions are {{{x = (-1+i*sqrt(47))/(6)}}} or {{{x = (-1-i*sqrt(47))/(6)}}} 



Note: {{{i = sqrt(-1)}}}


If you have not learned about complex numbers yet, then you can say "there are no real solutions"