Question 513533


{{{x^2-4x+7=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-4x+7}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-4}}}, and {{{C=7}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-4) +- sqrt( (-4)^2-4(1)(7) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-4}}}, and {{{C=7}}}



{{{x = (4 +- sqrt( (-4)^2-4(1)(7) ))/(2(1))}}} Negate {{{-4}}} to get {{{4}}}. 



{{{x = (4 +- sqrt( 16-4(1)(7) ))/(2(1))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16-28 ))/(2(1))}}} Multiply {{{4(1)(7)}}} to get {{{28}}}



{{{x = (4 +- sqrt( -12 ))/(2(1))}}} Subtract {{{28}}} from {{{16}}} to get {{{-12}}}



{{{x = (4 +- sqrt( -12 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (4 +- 2i*sqrt(3))/(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 = (4)/(2) +- (2i*sqrt(3))/(2)}}} Break up the fraction.  



{{{x = 2 +- sqrt(3)*i}}} Reduce.  



{{{x = 2+sqrt(3)*i}}} or {{{x = 2-sqrt(3)*i}}} Break up the expression.  



So the solutions are {{{x = 2+sqrt(3)*i}}} or {{{x = 2-sqrt(3)*i}}}