Question 551123


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



Notice that the quadratic {{{x^2+5x+6}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=5}}}, and {{{C=6}}}



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)(6) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=5}}}, and {{{C=6}}}



{{{x = (-5 +- sqrt( 25-4(1)(6) ))/(2(1))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25-24 ))/(2(1))}}} Multiply {{{4(1)(6)}}} to get {{{24}}}



{{{x = (-5 +- sqrt( 1 ))/(2(1))}}} Subtract {{{24}}} from {{{25}}} to get {{{1}}}



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



{{{x = (-5 +- 1)/(2)}}} Take the square root of {{{1}}} to get {{{1}}}. 



{{{x = (-5 + 1)/(2)}}} or {{{x = (-5 - 1)/(2)}}} Break up the expression. 



{{{x = (-4)/(2)}}} or {{{x =  (-6)/(2)}}} Combine like terms. 



{{{x = -2}}} or {{{x = -3}}} Simplify. 



So the solutions are {{{x = -2}}} or {{{x = -3}}}