Question 173543
To find the x-intercepts, simply set the right side equal to zero to get




{{{x^2+5x+1=0}}} 



Notice we have a quadratic equation in the form of {{{ax^2+bx+c=0}}} where {{{a=1}}}, {{{b=5}}}, 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 = (-(5) +- sqrt( (5)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=5}}}, and {{{c=1}}}



{{{x = (-5 +- sqrt( 25-4(1)(1) ))/(2(1))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{x = (-5 +- sqrt( 21 ))/(2(1))}}} Subtract {{{4}}} from {{{25}}} to get {{{21}}}



{{{x = (-5 +- sqrt( 21 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-5+sqrt(21))/(2)}}} or {{{x = (-5-sqrt(21))/(2)}}} Break up the expression.  



So the answers are {{{x = (-5+sqrt(21))/(2)}}} or {{{x = (-5-sqrt(21))/(2)}}} 



which approximate to {{{x=-0.209}}} or {{{x=-4.791}}} 




So the x-intercepts are approximately (-0.209,0) and (-4.791,0)