Question 151058

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



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



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



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



{{{x = (1 +- sqrt( -79 ))/(2(1))}}} Subtract {{{80}}} from {{{1}}} to get {{{-79}}}



{{{x = (1 +- sqrt( -79 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- i*sqrt(79))/(2)}}} Simplify the square root  



{{{x = (1+i*sqrt(79))/(2)}}} or {{{x = (1-i*sqrt(79))/(2)}}} Break up the expression.  



So our answers are {{{x = (1+i*sqrt(79))/(2)}}} or {{{x = (1-i*sqrt(79))/(2)}}} 



which approximate to {{{x=0.5+4.444*i}}} or {{{x=0.5-4.444*i}}}