Question 265119


{{{4t^2+9=12t}}} Start with the given equation.



{{{4t^2-12t+9=0}}} Subtract 12t from both sides.



Notice that the quadratic {{{4t^2-12t+9}}} is in the form of {{{At^2+Bt+C}}} where {{{A=4}}}, {{{B=-12}}}, and {{{C=9}}}



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



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



{{{t = (-(-12) +- sqrt( (-12)^2-4(4)(9) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-12}}}, and {{{C=9}}}



{{{t = (12 +- sqrt( (-12)^2-4(4)(9) ))/(2(4))}}} Negate {{{-12}}} to get {{{12}}}. 



{{{t = (12 +- sqrt( 144-4(4)(9) ))/(2(4))}}} Square {{{-12}}} to get {{{144}}}. 



{{{t = (12 +- sqrt( 144-144 ))/(2(4))}}} Multiply {{{4(4)(9)}}} to get {{{144}}}



{{{t = (12 +- sqrt( 0 ))/(2(4))}}} Subtract {{{144}}} from {{{144}}} to get {{{0}}}



{{{t = (12 +- sqrt( 0 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{t = (12 +- 0)/(8)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{t = (12 + 0)/(8)}}} or {{{t = (12 - 0)/(8)}}} Break up the expression. 



{{{t = (12)/(8)}}} or {{{t =  (12)/(8)}}} Combine like terms. 



{{{t = 3/2}}} or {{{t = 3/2}}} Simplify. 



So the solution is {{{t = 3/2}}}