how to solve one variable in non linear equation?

In my coding I couldn't get the value for t1.  I get the answer in cubic equation of another variable z. But I didn't  use z anywhere in my codind.  Anyone help and tell how to get the answer for t1.
clc
clear all
T=12; u1=4;u2=8;a=30;b=5;a2=100;A=500;c2=10;c3=12;c4=8;D0=115; b2=0.2;
a=0.01;d=0.2;m=0.5;k1=0.5;k0=1;k2=2;
syms t1  
c1=5;
h=(1./T).*(c1.*((b-a)-b.*t1-(a+b.*(1+m)./(1+m-t1)))+c2.*(b.*t1-(a+b.*(1+m)).*(t1./(1+m-t1)))+c3.*((a./k1)-((b./k1.^2).*(1-k1.*T)).*(((-u1.*k1)./(1+k1.*(t1-T))+1)-((1-k1.*T)./(1+k1.*(t1-T)))+(b.*k0.*(t1-u1)./k1)-(a-(b.*k0.*(1-k1.*T)./k1)).*((u2-u1)./(1+k1.*(t1-T)))+(k0.*D0./k1).*((k1.*(u1-u2)./(1+k1.*(t1-T)))-(a-(b.*k0./k1).*(1-k1.*T)).*((T-k2)./(1+k1.*(t1-T)))+(b.*k0./k1).*(u2-T)-(1-k1.*T).*((T+u2)./(1+k1.*(t1-T)))+c4.*(-a2+b2.*t1+k0.*(a-(b./k1).*(1-k1.*T).*(1./(1+k1.*(t1-T))+(b./k1))))))));
t1=solve(h)
disp(t1)

Answer 1

[Torsten]

T=12; u1=400;u2=8;a=30;b=5;a2=100;A=500;c2=10;c3=12;c4=8;D0=115; b2=0.2;
a=-1.1;d=11.2;m=-0.5;k1=1.5;k0=1.1;k2=2;
syms t1  
c1=5;
h= (1./T).*(c1.*((b-a)-b.*t1-(a+b.*(1+m)./(1+m-t1)))+c2.*(b.*t1-(a+b.*(1+m)).*(t1./(1+m-t1)))+c3.*((a./k1)-((b./k1.^2).*(1-k1.*T)).*(((-u1.*k1)./(1+k1.*(t1-T))+1)-((1-k1.*T)./(1+k1.*(t1-T)))+(b.*k0.*(t1-u1)./k1)-(a-(b.*k0.*(1-k1.*T)./k1)).*((u2-u1)./(1+k1.*(t1-T)))+(k0.*D0./k1).*((k1.*(u1-u2)./(1+k1.*(t1-T)))-(a-(b.*k0./k1).*(1-k1.*T)).*((T-k2)./(1+k1.*(t1-T)))+(b.*k0./k1).*(u2-T)-(1-k1.*T).*((T+u2)./(1+k1.*(t1-T)))+c4.*(-a2+b2.*t1+k0.*(a-(b./k1).*(1-k1.*T).*(1./(1+k1.*(t1-T))+(b./k1))))))));
t1=vpasolve(h,t1,[0 1])

t1 = 

0.4999993236378365887378927054944

double(subs(h,t1))

ans = 2.2769e-24

[M.Rameswari Sudha]

Thank you so much. I got the answer.

Answer 2

[VBBV]

clc
clear all
T=12; u1=400;u2=8;a=30;b=5;a2=100;A=500;c2=10;c3=12;c4=8;D0=115; b2=0.2;
a=-1.1;d=11.2;m=-0.5;k1=1.5;k0=1.1;k2=2;
syms t1  
c1=5;
h=@(t1) (1./T).*(c1.*((b-a)-b.*t1-(a+b.*(1+m)./(1+m-t1)))+c2.*(b.*t1-(a+b.*(1+m)).*(t1./(1+m-t1)))+c3.*((a./k1)-((b./k1.^2).*(1-k1.*T)).*(((-u1.*k1)./(1+k1.*(t1-T))+1)-((1-k1.*T)./(1+k1.*(t1-T)))+(b.*k0.*(t1-u1)./k1)-(a-(b.*k0.*(1-k1.*T)./k1)).*((u2-u1)./(1+k1.*(t1-T)))+(k0.*D0./k1).*((k1.*(u1-u2)./(1+k1.*(t1-T)))-(a-(b.*k0./k1).*(1-k1.*T)).*((T-k2)./(1+k1.*(t1-T)))+(b.*k0./k1).*(u2-T)-(1-k1.*T).*((T+u2)./(1+k1.*(t1-T)))+c4.*(-a2+b2.*t1+k0.*(a-(b./k1).*(1-k1.*T).*(1./(1+k1.*(t1-T))+(b./k1))))))));
t1=fsolve(h,[0 1])

Equation solved, solver stalled.

fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared and the vector of function values
is near zero as measured by the value of the function tolerance.

t1 = 1×2

10.4643 10.4643

disp(t1)

   10.4643   10.4643

[VBBV]

Check with fsolve instead of solve

[M.Rameswari Sudha]

you take [0 1] for t1. but ans is 10.463. How?. I need the answer between 0 to 1.

[M.Rameswari Sudha]

I want to solve t1 from h. No need to find h value. Help me

[VBBV]

Read about fsolve for more info. Try with different initial values. I have shown an e.g.