Question 156376
{{{(2x-1)^2=x^2-2}}} Start with the given equation



{{{4x^2-4x+1=x^2-2}}} FOIL the left side



{{{4x^2-4x+1-x^2+2=0}}} Subtract {{{x^2}}} from both sides. Add {{{2}}} to both sides.



{{{3x^2-4x+3=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{b=-4}}}, and {{{c=3}}}



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(3)(3) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-4}}}, and {{{c=3}}}



{{{x = (4 +- sqrt( (-4)^2-4(3)(3) ))/(2(3))}}} Negate {{{-4}}} to get {{{4}}}. 



{{{x = (4 +- sqrt( 16-4(3)(3) ))/(2(3))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16-36 ))/(2(3))}}} Multiply {{{4(3)(3)}}} to get {{{36}}}



{{{x = (4 +- sqrt( -20 ))/(2(3))}}} Subtract {{{36}}} from {{{16}}} to get {{{-20}}}



{{{x = (4 +- sqrt( -20 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (4 +- 2i*sqrt(5))/(6)}}} 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+2i*sqrt(5))/(6)}}} or {{{x = (4-2i*sqrt(5))/(6)}}} Break up the expression.  



{{{x = (2+i*sqrt(5))/(3)}}} or {{{x = (2-i*sqrt(5))/(3)}}} Reduce



So the answers are {{{x = (2+i*sqrt(5))/(3)}}} or {{{x = (2-i*sqrt(5))/(3)}}} 



which approximate to {{{x=0.667+0.745*i}}} or {{{x=0.667-0.745*i}}}