Question

Solve the equation by identifying the quadratic form. 6(t-1)^(2)-19(t-1)=-3

Answer 1

6t^2-31t+28=0

Explanation:

6(t-1)^2-19(t-1)=-3

6(t^2-27+1)-19t+19=-3

6t^2-31t+25+3

6t^2-31t+28=0

(6t-7)(t-4)=0