Question 592287


{{{x^2+9x+20=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=9}}}, and {{{c=20}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(9) +- sqrt( (9)^2-4(1)(20) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=9}}}, and {{{c=20}}}



{{{x = (-9 +- sqrt( 81-4(1)(20) ))/(2(1))}}} Square {{{9}}} to get {{{81}}}. 



{{{x = (-9 +- sqrt( 81-80 ))/(2(1))}}} Multiply {{{4(1)(20)}}} to get {{{80}}}



{{{x = (-9 +- sqrt( 1 ))/(2(1))}}} Subtract {{{80}}} from {{{81}}} to get {{{1}}}



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



{{{x = (-9 +- 1)/(2)}}} Take the square root of {{{1}}} to get {{{1}}}. 



{{{x = (-9 + 1)/(2)}}} or {{{x = (-9 - 1)/(2)}}} Break up the expression. 



{{{x = (-8)/(2)}}} or {{{x =  (-10)/(2)}}} Combine like terms. 



{{{x = -4}}} or {{{x = -5}}} Simplify. 



So the answers are {{{x = -4}}} or {{{x = -5}}}