Question 236647

{{{x^2-x+49=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-x+49}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=49}}}



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)(49) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=49}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(49) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(49) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-196 ))/(2(1))}}} Multiply {{{4(1)(49)}}} to get {{{196}}}



{{{x = (1 +- sqrt( -195 ))/(2(1))}}} Subtract {{{196}}} from {{{1}}} to get {{{-195}}}



{{{x = (1 +- sqrt( -195 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- i*sqrt(195))/(2)}}} Simplify the square root  



{{{x = (1+i*sqrt(195))/(2)}}} or {{{x = (1-i*sqrt(195))/(2)}}} Break up the expression.  



So the solutions are {{{x = (1+i*sqrt(195))/(2)}}} or {{{x = (1-i*sqrt(195))/(2)}}}