Question 387322
{{{t^2+10=6t}}} Start with the given equation.



{{{t^2+10-6t=0}}} Subtract 6t from both sides.



{{{t^2-6t+10=0}}} Rearrange the terms.



Notice that the quadratic {{{t^2-6t+10}}} is in the form of {{{At^2+Bt+C}}} where {{{A=1}}}, {{{B=-6}}}, and {{{C=10}}}



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



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



{{{t = (-(-6) +- sqrt( (-6)^2-4(1)(10) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-6}}}, and {{{C=10}}}



{{{t = (6 +- sqrt( (-6)^2-4(1)(10) ))/(2(1))}}} Negate {{{-6}}} to get {{{6}}}. 



{{{t = (6 +- sqrt( 36-4(1)(10) ))/(2(1))}}} Square {{{-6}}} to get {{{36}}}. 



{{{t = (6 +- sqrt( 36-40 ))/(2(1))}}} Multiply {{{4(1)(10)}}} to get {{{40}}}



{{{t = (6 +- sqrt( -4 ))/(2(1))}}} Subtract {{{40}}} from {{{36}}} to get {{{-4}}}



{{{t = (6 +- sqrt( -4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{t = (6 +- 2*i)/(2)}}} Take the square root of {{{-4}}} to get {{{2*i}}}. 



{{{t = (6 + 2*i)/(2)}}} or {{{t = (6 - 2*i)/(2)}}} Break up the expression. 



{{{t = (6)/(2) + (2*i)/(2)}}} or {{{t =  (6)/(2) - (2*i)/(2)}}} Break up the fraction for each case. 



{{{t = 3+i}}} or {{{t =  3-i}}} Reduce. 



So the solutions are {{{t = 3+i}}} or {{{t = 3-i}}} 



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