Explanation:
Given equation is 2t/(t-5) + 1/(t-3) = 3
=> [2t(t-3)+1(t-5)]/(t-5)(t-3) = 3
=> (2t²-6t+t-5)/(t²-5t-3t+15) = 3
=> (2t²-5t-5)/(t²-8t+15) = 3
=> 2t²-5t-5 = 3(t²-8t+15)
=> 2t²-5t-5 = 3t²-24t+45
=>3t²-24t+45-2t²+5t+5 = 0
=> t²-19t+50 = 0
The standard form is t²-19t+50 = 0