Question 150465
I'm assuming that you mean {{{x^2+3x+1=0}}}



{{{x^2+3x+1=0}}} Start with the given equation.



Let's use the quadratic formula to solve for x



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



{{{x = (-(3) +- sqrt( (3)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=3}}}, and {{{c=1}}}



{{{x = (-3 +- sqrt( 9-4(1)(1) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{x = (-3 +- sqrt( 9-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{x = (-3 +- sqrt( 5 ))/(2(1))}}} Subtract {{{4}}} from {{{9}}} to get {{{5}}}



{{{x = (-3 +- sqrt( 5 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-3+sqrt(5))/(2)}}} or {{{x = (-3-sqrt(5))/(2)}}} Break up the expression.  



So our answers are {{{x = (-3+sqrt(5))/(2)}}} or {{{x = (-3-sqrt(5))/(2)}}} 



which approximate to {{{x=-0.382}}} or {{{x=-2.618}}}