the difference bt the roots of equaion (t+3)2+9(t-1)+k=0 is 5..then value of k is

(t+3)² + 9(t-1) + k=0

=> t² + 6t + 9 + 9t - 9 + k=0

=>t² + 15t + k=0

let the roots of the above quad.eq be h and k.

sum of roots h+k = -b/a

                  = -15/1

                  = -15  . .......(eq 1)

difference of roots h - k = 5 (given).......(eq 2)

solving eq1 and eq2,

we get h=-5 and k=-10

product of roots h*k = c/a

=>          (-5 X -10) = k/1

=> k=50

Approved

I will present another method of solving the problem

(difference between roots)²=(sum of roots)²-4(product of roots)