Question 184812


{{{x^2+5x+4=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=5}}}, and {{{c=4}}}



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)(4) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=5}}}, and {{{c=4}}}



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



{{{x = (-5 +- sqrt( 25-16 ))/(2(1))}}} Multiply {{{4(1)(4)}}} to get {{{16}}}



{{{x = (-5 +- sqrt( 9 ))/(2(1))}}} Subtract {{{16}}} from {{{25}}} to get {{{9}}}



{{{x = (-5 +- sqrt( 9 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-5 +- 3)/(2)}}} Take the square root of {{{9}}} to get {{{3}}}. 



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



{{{x = (-2)/(2)}}} or {{{x =  (-8)/(2)}}} Combine like terms. 



{{{x = -1}}} or {{{x = -4}}} Simplify. 



So the answers are {{{x = -1}}} or {{{x = -4}}}