Question 460095


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



{{{2x^2-4x+2x=0}}} Get every term to the left side.



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



Notice that the quadratic {{{2x^2-2x}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=-2}}}, and {{{C=0}}}



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



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



{{{x = (-(-2) +- sqrt( (-2)^2-4(2)(0) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=-2}}}, and {{{C=0}}}



{{{x = (2 +- sqrt( (-2)^2-4(2)(0) ))/(2(2))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{x = (2 +- sqrt( 4-4(2)(0) ))/(2(2))}}} Square {{{-2}}} to get {{{4}}}. 



{{{x = (2 +- sqrt( 4-0 ))/(2(2))}}} Multiply {{{4(2)(0)}}} to get {{{0}}}



{{{x = (2 +- sqrt( 4 ))/(2(2))}}} Subtract {{{0}}} from {{{4}}} to get {{{4}}}



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



{{{x = (2 +- 2)/(4)}}} Take the square root of {{{4}}} to get {{{2}}}. 



{{{x = (2 + 2)/(4)}}} or {{{x = (2 - 2)/(4)}}} Break up the expression. 



{{{x = (4)/(4)}}} or {{{x =  (0)/(4)}}} Combine like terms. 



{{{x = 1}}} or {{{x = 0}}} Simplify. 



So the solutions are {{{x = 1}}} or {{{x = 0}}}