Question 601460


{{{2x^2-12x+10=0}}} Start with the given equation.



Notice that the quadratic {{{2x^2-12x+10}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=-12}}}, 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 = (-(-12) +- sqrt( (-12)^2-4(2)(10) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=-12}}}, and {{{C=10}}}



{{{x = (12 +- sqrt( (-12)^2-4(2)(10) ))/(2(2))}}} Negate {{{-12}}} to get {{{12}}}. 



{{{x = (12 +- sqrt( 144-4(2)(10) ))/(2(2))}}} Square {{{-12}}} to get {{{144}}}. 



{{{x = (12 +- sqrt( 144-80 ))/(2(2))}}} Multiply {{{4(2)(10)}}} to get {{{80}}}



{{{x = (12 +- sqrt( 64 ))/(2(2))}}} Subtract {{{80}}} from {{{144}}} to get {{{64}}}



{{{x = (12 +- sqrt( 64 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (12 +- 8)/(4)}}} Take the square root of {{{64}}} to get {{{8}}}. 



{{{x = (12 + 8)/(4)}}} or {{{x = (12 - 8)/(4)}}} Break up the expression. 



{{{x = (20)/(4)}}} or {{{x =  (4)/(4)}}} Combine like terms. 



{{{x = 5}}} or {{{x = 1}}} Simplify. 



So the solutions are {{{x = 5}}} or {{{x = 1}}}