Question 275252
The zeros are the x values that satisfy the equation {{{x^2+6x+5=0}}}



{{{x^2+6x+5=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+6x+5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=6}}}, 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 = (-(6) +- sqrt( (6)^2-4(1)(5) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=6}}}, and {{{C=5}}}



{{{x = (-6 +- sqrt( 36-4(1)(5) ))/(2(1))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36-20 ))/(2(1))}}} Multiply {{{4(1)(5)}}} to get {{{20}}}



{{{x = (-6 +- sqrt( 16 ))/(2(1))}}} Subtract {{{20}}} from {{{36}}} to get {{{16}}}



{{{x = (-6 +- sqrt( 16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-6 +- 4)/(2)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{x = (-6 + 4)/(2)}}} or {{{x = (-6 - 4)/(2)}}} Break up the expression. 



{{{x = (-2)/(2)}}} or {{{x =  (-10)/(2)}}} Combine like terms. 



{{{x = -1}}} or {{{x = -5}}} Simplify. 



So the zeros are {{{x = -1}}} or {{{x = -5}}}