Question 561375


{{{5t^2-27t+22=0}}} Start with the given equation.



Notice that the quadratic {{{5t^2-27t+22}}} is in the form of {{{At^2+Bt+C}}} where {{{A=5}}}, {{{B=-27}}}, and {{{C=22}}}



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



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



{{{t = (-(-27) +- sqrt( (-27)^2-4(5)(22) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=-27}}}, and {{{C=22}}}



{{{t = (27 +- sqrt( (-27)^2-4(5)(22) ))/(2(5))}}} Negate {{{-27}}} to get {{{27}}}. 



{{{t = (27 +- sqrt( 729-4(5)(22) ))/(2(5))}}} Square {{{-27}}} to get {{{729}}}. 



{{{t = (27 +- sqrt( 729-440 ))/(2(5))}}} Multiply {{{4(5)(22)}}} to get {{{440}}}



{{{t = (27 +- sqrt( 289 ))/(2(5))}}} Subtract {{{440}}} from {{{729}}} to get {{{289}}}



{{{t = (27 +- sqrt( 289 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{t = (27 +- 17)/(10)}}} Take the square root of {{{289}}} to get {{{17}}}. 



{{{t = (27 + 17)/(10)}}} or {{{t = (27 - 17)/(10)}}} Break up the expression. 



{{{t = (44)/(10)}}} or {{{t =  (10)/(10)}}} Combine like terms. 



{{{t = 22/5}}} or {{{t = 1}}} Simplify. 



So the solutions are {{{t = 22/5}}} or {{{t = 1}}} 



Note: 22/5 = 4.4



So you have the correct answers.

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