What is t such as the definite integral of 4x^3+16 is 31 if the limits of integration are t and t+1 ?

Let f(x)= 4x^3 + 16

Let F(x)
= Int f(x)

Then we are given that F(t+1) - F(t) =
31

Let us integrate
f(x).

==> F(x)= Int 4x^3 +16 
dx

==> F(x)= x^4 +
16x

==> F(t+1)= (t+1)^4 +
14(t+1)

==> F(t) = t^4 +
16t

==> F(t+1) - F(t) =
31

==> (t+1)^4 + 16(t+1) - t^4 -16t =
31

==> (t+1)^4 + 16t + 16 - t^4 - 16t =
31

Reduce similar.

==>
t^4 + 4t^3+ 6t^2 +4t + 1 + 16t +16- t^4 - 16t =
31

==> 4t^3 + 6t^2 + 4t +1 + 16 =
31

==> 4t^3 + 6t^2 + 4t - 14 =
0

==> (t-1) ( 4t^2+10t+14) =
0

==> 2(t-1)(2t^2 +5t +7)=
0

==> t1= 1

==>
t2= ( -5 + sqrt(25-4*2*7) /2*2

            = (-5 + sqrt-31
) / 4

             = (-5/4) + sqrt31)/4 *
i

==> t2= (-5/4) - sqrt31/4
*i

Then we have 3 values for
t.

==> t= { 1, (-5/4)+sqrt31/4 *i )  
, (-5/4 - sqrt31/4 *i) }