Question 161413


{{{6x^2+x+4=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=6}}}, {{{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(6)(4) ))/(2(6))}}} Plug in  {{{a=6}}}, {{{b=1}}}, and {{{c=4}}}



{{{x = (-1 +- sqrt( 1-4(6)(4) ))/(2(6))}}} Square {{{1}}} to get {{{1}}}. 



{{{x = (-1 +- sqrt( 1-96 ))/(2(6))}}} Multiply {{{4(6)(4)}}} to get {{{96}}}



{{{x = (-1 +- sqrt( -95 ))/(2(6))}}} Subtract {{{96}}} from {{{1}}} to get {{{-95}}}



{{{x = (-1 +- sqrt( -95 ))/(12)}}} Multiply {{{2}}} and {{{6}}} to get {{{12}}}. 



{{{x = (-1 +- i*sqrt(95))/(12)}}} Simplify the square root  



{{{x = (-1+i*sqrt(95))/(12)}}} or {{{x = (-1-i*sqrt(95))/(12)}}} Break up the expression.  



So the answers are {{{x = (-1+i*sqrt(95))/(12)}}} or {{{x = (-1-i*sqrt(95))/(12)}}}