How to find L from the given code?

clc

close all

lags = 0.6;

k1 = 3;

k_1 = 1;

k2 = 2.5;

k3 = 1;

k_3 = 1;

k4 = 2;

k5 = 1;

E1 = 1;

E2 = 2;

K1 = (k_1+k2)/k1;

K2 = (k_3+k4)/k3;

k3*E2*(L^2+L*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2)*(k1*k2*E1+(L+k1*K1+k1*E1)*(L-k5*E1*e^(-L*lags))) = 0;

Answer 1

[Torsten]

lags = 0.6;
k1 = 3;
k_1 = 1;
k2 = 2.5;
k3 = 1;
k_3 = 1;
k4 = 2;
k5 = 1;
E1 = 1;
E2 = 2;
K1 = (k_1+k2)/k1;
K2 = (k_3+k4)/k3;
fun = @(L) k3*E2*(L.^2+L.*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2).*(k1*k2*E1+(L+k1*K1+k1*E1).*(L-k5*E1*exp(-L*lags)));
L0 = -1.5;
L1 = fsolve(fun,L0)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

L1 = -1.6620

L0 = -2.5;
L2 = fsolve(fun,L0)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

L2 = -2.4088

plot((-2.5:0.1:-1),fun(-2.5:0.1:-1))

[Akhtar Jan]

Thanks alot sir

Answer 2

[Walter Roberson]

syms L
e = exp(sym(1))

e = 

e

lags = 0.6;
k1 = 3;
k_1 = 1;
k2 = 2.5;
k3 = 1;
k_3 = 1;
k4 = 2;
k5 = 1;
E1 = 1;
E2 = 2;
K1 = (k_1+k2)/k1;
K2 = (k_3+k4)/k3;
eqn = k3*E2*(L^2+L*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2)*(k1*k2*E1+(L+k1*K1+k1*E1)*(L-k5*E1*e^(-L*lags)))

eqn = 

vpasolve(eqn)

ans = 

There is a second solution near -2.3

[Akhtar Jan]

thank you sir

[Akhtar Jan]

this one is also helpfull