Question 333720


{{{t^2=9t}}} Start with the given equation.



{{{t^2-9t=0}}} Subtract 9t from both sides.



Notice that the quadratic {{{t^2-9t}}} is in the form of {{{At^2+Bt+C}}} where {{{A=1}}}, {{{B=-9}}}, and {{{C=0}}}



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



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



{{{t = (-(-9) +- sqrt( (-9)^2-4(1)(0) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-9}}}, and {{{C=0}}}



{{{t = (9 +- sqrt( (-9)^2-4(1)(0) ))/(2(1))}}} Negate {{{-9}}} to get {{{9}}}. 



{{{t = (9 +- sqrt( 81-4(1)(0) ))/(2(1))}}} Square {{{-9}}} to get {{{81}}}. 



{{{t = (9 +- sqrt( 81-0 ))/(2(1))}}} Multiply {{{4(1)(0)}}} to get {{{0}}}



{{{t = (9 +- sqrt( 81 ))/(2(1))}}} Subtract {{{0}}} from {{{81}}} to get {{{81}}}



{{{t = (9 +- sqrt( 81 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{t = (9 +- 9)/(2)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{t = (9 + 9)/(2)}}} or {{{t = (9 - 9)/(2)}}} Break up the expression. 



{{{t = (18)/(2)}}} or {{{t =  (0)/(2)}}} Combine like terms. 



{{{t = 9}}} or {{{t = 0}}} Simplify. 



So the solutions are {{{t = 9}}} or {{{t = 0}}}