Newton's Method - matlab

Question

I am trying to find out the root of an equation constraint using Newton's Method (open to any other method). The equation constraint is dependent on the roots of a quadratic equation which has one unknown term. The equation descriptions (quadratic and constraint) are shown below. The roots of the quadratic equation are assumed to be p1 and p2.
x^2 + x*(c1*unknown/2 + C1*c2)/(c1*c2*c3*unknown/2) + 1/(c1*c2*c3*u/2) = 0
with constraint
(1/(p2-p1))*(exp(-0.3*p2) - 1)*(c1*c2 - p2) - (exp(-0.3*p1) - 1)*(c1*c2 - p1) - 0.04 = 0
I am wondering if there are any other approximation methods to solve this problem if Newton's Method is not going to do so.
Matlab Code
p0=10*10^6;
Cc=0.65*10^-6;Rp=100*10^3;Cp=55*10^-9; z1=1/(Cp*Rp);
N = 100;error = 0.02;
syms 'x'
a = 1;
b = ((Cc*(x/2+Rp))/(Cc*Cp*Rp*x/2));
c = 1/(Cc*Cp*Rp*x/2);
poles1 = (-b + sqrt(b^2 - 4*a*c))/(2*a);
poles2 = (-b - sqrt(b^2 - 4*a*c))/(2*a);
p1_subterm = (exp(-0.3*poles1) - 1)*(z1 - poles1);
p2_subterm = (exp(-0.3*poles2) - 1)*(z1- poles2);
f(x) = (1/(poles2 - poles1))*p2_subterm - p2_subterm - 0.04;
df = diff(f);
while i <= N
p = p0-(f(p0)/df(p0))
if (abs(p - p0)/abs(p)) < error
fprintf('Solution is %f \n', double(p))
return
end
i = i + 1;
p0 = p;
end
fprintf('Solution did not converge within %d iterations \n', N)

Answer 1

I don't think your code for f(x) corresponds to the equation listed above: In the MATLAB code 1/(poles2-poles1) multiplies both the exp((...)poles2) and the exp((...)poles1) terms and you have 1/(c1*c2) - poles1 instead of c1*c2 - poles1.
I have not checked through the poles1 and poles2 terms as I find the 10 sets of parenthesis intimidating. It would suggest making your code more readable by breaking up each of those equations in to a number of shorter, clearer lines.
Noting that we can multiply the quadratic by c1*c2*c3*unknown,
(c1*c2*c3*unknown)*x^2 + x*2*(c1*unknown/2 + C1*c2) + 2 = 0
then you can write the poles and f code as
% Here unknown is x
syms 'x'
a = c1*c2*c3*x;
b = c1*x + 2*c1*c2;
c = 2;
poles1 = (-b + sqrt(b^2 - 4*a*c))/(2*a);
poles2 = (-b - sqrt(b^2 - 4*a*c))/(2*a);
p1_subterm = (exp(-0.3*poles1) - 1)*(c1*c2 - poles1);
p2_subterm = (exp(-0.3*poles2) - 1)*(c1*c2 - poles2);
f(x) = (1/(poles2 - poles1))*p2_subterm - p2_subterm - 0.04;