Question 170864
I'm assuming that you want to solve {{{x^2+x-5=0}}}



{{{x^2+x-5=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=1}}}, and {{{c=-5}}}



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(1)(-5) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=1}}}, and {{{c=-5}}}



{{{x = (-1 +- sqrt( 1-4(1)(-5) ))/(2(1))}}} Square {{{1}}} to get {{{1}}}. 



{{{x = (-1 +- sqrt( 1--20 ))/(2(1))}}} Multiply {{{4(1)(-5)}}} to get {{{-20}}}



{{{x = (-1 +- sqrt( 1+20 ))/(2(1))}}} Rewrite {{{sqrt(1--20)}}} as {{{sqrt(1+20)}}}



{{{x = (-1 +- sqrt( 21 ))/(2(1))}}} Add {{{1}}} to {{{20}}} to get {{{21}}}



{{{x = (-1 +- sqrt( 21 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-1+sqrt(21))/(2)}}} or {{{x = (-1-sqrt(21))/(2)}}} Break up the expression.  



So the answers are {{{x = (-1+sqrt(21))/(2)}}} or {{{x = (-1-sqrt(21))/(2)}}} 



which approximate to {{{x=1.791}}} or {{{x=-2.791}}}