There are different ways to solve this problem. For example, it is a sequence of digits after the decimal point in the expansion of the fraction  248590/1369863  to decimal fraction:

evalf[48](248590/1369863);

                         0.181470701814707018147070181470701814707018147070

restart;

ys := [1.8*a+.4000000000*b+103.9*c-.5000000000*e-102.6*g = 0, .5000000000*b+102.6*c+.3*d+.3500000000*e-102.1*g = 0, -3.3*a-1.050000000*b+102.4*c+.7*d+.2500000000*e-105.1*g = 0, -2.5*a+1.450000000*b+102.6*c+3.3*d+1.050000000*e-102.4*g = 0, -1.6*a+1.100000000*b+105.8*c+.3*d+1.750000000*e-105.5*g = 0, -.7*a-.2500000000*b+105.1*c-.2*d+.2000000000*e-105.7*g = 0, -.3*a-.3500000000*b+102.1*c+2.5*d-1.450000000*e-102.6*g = 0, .2*a-.2000000000*b+105.7*c+1.6*d-1.100000000*e-105.8*g = 0]:

B := [a, b, c, d, e, g]:

A:=Matrix([seq([seq(coeff(lhs(ys[i]), B[j]), j=1..nops(B))], i=1..nops(ys))]);

Your system has infinitely many solutions, depending on two parameters  a  and  g .

The decision with checking:

sys:=[1.8*a+.4000000000*b+103.9*c-.5000000000*e-102.6*g = 0, .5000000000*b+102.6*c+.3*d+.3500000000*e-102.1*g = 0, -.3*a-.3500000000*b+102.1*c+2.5*d-1.450000000*e-102.6*g = 0, -2.5*a+1.450000000*b+102.6*c+3.3*d+1.050000000*e-102.4*g = 0, -3.3*a-1.050000000*b+102.4*c+.7*d+.2500000000*e-105.1*g = 0, -.7*a-.2500000000*b+105.1*c-.2*d+.2000000000*e-105.7*g = 0]:

sol:=solve(sys);

eval(sys, sol);

isolve(24*s+33*t=9);

isolve(6*x+10*y+15*z=7);

_Z1, _Z2  -  any integers

restart;

for N while is((1/6)*Pi^2-sum(1/n^2, n = 1 .. N) >= 0.1e-2) do

od;

N; 

                              1000

First - collect, then - exp:

restart;

a0:=I*(m*x+mp*x-omega[m]*t-omega[mp]*t);

a1:=exp(collect(a0, [x,t]));

     # or

a2:=exp(a0);

applyop(collect, 1, a2, [x, t]); 

Or use  a more powerful  algsubs  command:

w:=x+2+sin(x+2);

algsubs(x+2=u, w);

          w := x+2+sin(x+2)
                  u+sin(u)

PP:=proc(S::set)

local It, Sp, s;

if S={} then return {{}} fi;

It:=proc(Sp, s)

{op(Sp), seq(Sp[i] union {s}, i=1..nops(Sp))}

end;

Sp:={{}};

   for s in S do

   Sp:=It(Sp, s)

   od;

Sp;

end;

Example:

PP({a,b,c,d,e});

{{}, {a}, {b}, {c}, {d}, {e}, {a, b}, {a, c}, {a, d}, {a, e}, {b, c}, {b, d}, {b, e}, {c, d}, {c, e}, {d, e}, {a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}, {a, d, e}, {b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}, {a, b, c, d}, {a, b, c, e}, {a, b, d, e}, {a, c, d, e}, {b, c, d, e}, {a, b, c, d, e}}

X:=Array([seq([x, x^2, 8-4*x], x=0..2.9, 0.1)]);

plot([seq(X[i,2],i=1..30)], [seq(X[i,1],i=1..30)]);

             # or

plot([seq(X[i,3],i=1..30)], [seq(X[i,1],i=1..30)]);

See  http://www.physics.arizona.edu/~restrepo/475A/Notes/sourcea-/node23.html

I have reduced the equation by  x^2  removing the trivial root  x=0.

plots[implicitplot](x+omega^2*x^3-omega=0, omega=-2..10, x=-1..1, thickness=2, numpoints=20000);

It is easy to guess the rule: if  n  is even then  F(n)=F(n-1)*3  else  F(n)=F(n-1)+9  so

F:=proc(n)

option remember;

if n=1 then 3 else

((-1)^n+1)/2*3*F(n-1)+(1-(-1)^n)/2*(F(n-1)+9) fi;

end:

seq(F(n), n=1..15);

                3, 9, 18, 54, 63, 189, 198, 594, 603, 1809, 1818, 5454, 5463, 16389, 16398

Enequality  n^2<=2^n  is true for all integer  n>=4 .

Two variants:

restart:

is(n^2<=2^n) assuming n::integer, n>=4;

restart:

assume(n::integer, n>=4):

is(n^2<=2^n);

                       true

                       true

l := 89: h := 49: d:= 55: beta1 := 11.5*Pi/180:

A := l*sin(beta1): B := l*cos(beta1): C := (h + 0.5*d)*sin(beta1) - 0.5*d*tan(beta1):

E := (h + 0.5*d)*cos(beta1) - 0.5*d:

RootFinding[Analytic](A*sin(alpha)*cos(alpha) + B*sin(2*alpha) - C*cos(alpha) - E*sin(alpha) = 0, alpha, re=0..2*Pi, im=-1..1);

evalf(map(convert,[%], degrees));

There is no  33 degrees or  22 degrees  in this list .

Your formula for the width of the wall of the players can be simplified since

sqrt(6400*x^4+12800*x^2*y^2+6400*y^4) = 80*(x^2+y^2)

and needs some refinement. This formula is true if it is assumed that the width of a football goal is 8 meters, and the distance from the point  (x, y)  to the wall is 10 meters. In fact, the gate width should be equal to 7.32 meters, and the distance to the wall  9 meters. 

With these adjustments more accurate formula:

z = (65.88*(x^2+y^2))*y/abs((x^2+y^2-3.66*x)*(3.66*x+x^2+y^2))

We construct a plot of this function, taking into account that the domain is a football field minus the penalty area:

restart;

z := unapply(piecewise(abs(x)<20.15 and y>0 and y<16.5, undefined, (65.88*(x^2+y^2)*y)/abs((x^2+y^2-3.66*x)*(3.66*x+x^2+y^2))), x, y):

plot3d(z, -34..34, 0..40, view=[-34..34, 0..40, 0..4.5], style=surface, axes=normal, numpoints=10000, orientation=[-40, 75], lightmodel=light4, labels=[x, y, z]);