Question: solved numerically a system of nonlinear algebraic equations

Hi everyone:

How can I solve numerically the system of nonlinear algebraic equations by Newton’s method?

eq1:= (1/2)*x[0]*sqrt(3)-(1/2)*x[1]*sqrt(3) = ((1/2)*x[0]*(t+(1/3)*sqrt(3))*sqrt(3)-(1/2)*x[1]*(t-(1/3)*sqrt(3))*sqrt(3))*(1-(1/6)*y[0]*(t+(1/3)*sqrt(3))*sqrt(3)+(1/6)*y[1]*(t-(1/3)*sqrt(3))*sqrt(3)-(1/8)*y[0]*(5*sqrt(3)*(1/12)-1/4+t)*sqrt(3)+(1/8)*y[1]*(-(1/4)*sqrt(3)-1/4+t)*sqrt(3)-(1/8)*y[0]*((1/4)*sqrt(3)-1/4+t)*sqrt(3)+(1/8)*y[1]*(-5*sqrt(3)*(1/12)-1/4+t)*sqrt(3))-5*t^3*(1/2)+49*t^2*(1/12)+17*t*(1/12)-23/6;
eq2:= (1/2)*y[0]*sqrt(3)-(1/2)*y[1]*sqrt(3) = ((1/2)*y[0]*(t+(1/3)*sqrt(3))*sqrt(3)-(1/2)*y[1]*(t-(1/3)*sqrt(3))*sqrt(3))*(-2+(1/2)*x[0]*(t+(1/3)*sqrt(3))*sqrt(3)-(1/2)*x[1]*(t-(1/3)*sqrt(3))*sqrt(3)+(1/4)*(-(1/12)*sqrt(3)-3/4)*((1/2)*x[0]*(5*sqrt(3)*(1/12)-1/4+t)*sqrt(3)-(1/2)*x[1]*(-(1/4)*sqrt(3)-1/4+t)*sqrt(3))+(1/4)*((1/12)*sqrt(3)-3/4)*((1/2)*x[0]*((1/4)*sqrt(3)-1/4+t)*sqrt(3)-(1/2)*x[1]*(-5*sqrt(3)*(1/12)-1/4+t)*sqrt(3)))+15*t^3*(1/8)-(1/4)*t^2+3*t*(1/8)-1;
eq3:=(1/2)*x[0]+(1/2)*x[1] = 1;
eq4:=(1/2)*y[0]+(1/2)*y[1] = 0;