Question 157994


{{{x^2+20x+101=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=20}}}, and {{{c=101}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(20) +- sqrt( (20)^2-4(1)(101) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=20}}}, and {{{c=101}}}



{{{x = (-20 +- sqrt( 400-4(1)(101) ))/(2(1))}}} Square {{{20}}} to get {{{400}}}. 



{{{x = (-20 +- sqrt( 400-404 ))/(2(1))}}} Multiply {{{4(1)(101)}}} to get {{{404}}}



{{{x = (-20 +- sqrt( -4 ))/(2(1))}}} Subtract {{{404}}} from {{{400}}} to get {{{-4}}}



{{{x = (-20 +- sqrt( -4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-20 +- 2*i)/(2)}}} Take the square root of {{{-4}}} to get {{{2*i}}}. 



{{{x = (-20 + 2*i)/(2)}}} or {{{x = (-20 - 2*i)/(2)}}} Break up the expression. 



{{{x = (-20)/(2) + (2*i)/(2)}}} or {{{x =  (-20)/(2) - (2*i)/(2)}}} Break up the fraction for each case. 



{{{x = -10+i}}} or {{{x =  -10-i}}} Reduce. 



So the answers are {{{x = -10+i}}} or {{{x =  -10-i}}}