Question 237737
{{{(t-5)^2=2(5-t)}}} Start with the given equation.



{{{t^2-10t+25=2(5-t)}}} FOIL



{{{t^2-10t+25=10-2t}}} Distribute



{{{t^2-10t+25-10+2t=0}}} Subtract 10 from both sides. Add 2t to both sides.



{{{t^2-8t+15=0}}} Combine like terms.



Notice that the quadratic {{{t^2-8t+15}}} is in the form of {{{At^2+Bt+C}}} where {{{A=1}}}, {{{B=-8}}}, and {{{C=15}}}



Let's use the quadratic formula to solve for "t":



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



{{{t = (-(-8) +- sqrt( (-8)^2-4(1)(15) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-8}}}, and {{{C=15}}}



{{{t = (8 +- sqrt( (-8)^2-4(1)(15) ))/(2(1))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{t = (8 +- sqrt( 64-4(1)(15) ))/(2(1))}}} Square {{{-8}}} to get {{{64}}}. 



{{{t = (8 +- sqrt( 64-60 ))/(2(1))}}} Multiply {{{4(1)(15)}}} to get {{{60}}}



{{{t = (8 +- sqrt( 4 ))/(2(1))}}} Subtract {{{60}}} from {{{64}}} to get {{{4}}}



{{{t = (8 +- sqrt( 4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{t = (8 +- 2)/(2)}}} Take the square root of {{{4}}} to get {{{2}}}. 



{{{t = (8 + 2)/(2)}}} or {{{t = (8 - 2)/(2)}}} Break up the expression. 



{{{t = (10)/(2)}}} or {{{t =  (6)/(2)}}} Combine like terms. 



{{{t = 5}}} or {{{t = 3}}} Simplify. 



So the solutions are {{{t = 5}}} or {{{t = 3}}}