Question 557710


{{{x^2-x=-10}}} Start with the given equation.



{{{x^2-x+10=0}}} Get every term to the left side.



Notice that the quadratic {{{x^2-x+10}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=10}}}



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)(10) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=10}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(10) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(10) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-40 ))/(2(1))}}} Multiply {{{4(1)(10)}}} to get {{{40}}}



{{{x = (1 +- sqrt( -39 ))/(2(1))}}} Subtract {{{40}}} from {{{1}}} to get {{{-39}}}



{{{x = (1 +- sqrt( -39 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- i*sqrt(39))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (1+i*sqrt(39))/(2)}}} or {{{x = (1-i*sqrt(39))/(2)}}} Break up the expression.  



So the solutions are {{{x = (1+i*sqrt(39))/(2)}}} or {{{x = (1-i*sqrt(39))/(2)}}} 



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