jmc

We have that $x^2=\frac{\sqrt{53}}{2}+\frac{3}{2}.$ Then $2x^2=\sqrt{53}+3,$ so $2x^2-3=\sqrt{53}.$ Squaring both sides, we get $$4x^4-12x^2+9=53,$$so $4x^4=12x^2+44.$ Then $x^4=3x^2+11.$

Since $x\ne 0,$ we can divide both sides of the given equation by $x^{40},$ to get $$x^{60}=2x^{58}+14x^{56}+11x^{54}-x^{10}+a{x}^6+b{x}^4+c.$$Now, $$\begin{array}{rl}{x}^{60}-2x^{58}-14x^{56}-11x^{54}& =x^{54}(x^6-2x^4-14x^2-11)\\ & =x^{54}((x^2-2)x^4-14x^2-11)\\ & =x^{54}((x^2-2)(3x^2+11)-14x^2-11)\\ & =x^{54}(3x^4-9x^2-33)\\ & =3x^{54}(x^4-3x^2-11)\\ & =0.\end{array}$$So, the equation reduces to $$x^{10}=a{x}^6+b{x}^4+c.$$We have that $$\begin{array}{rl}{x}^6& =x^2\cdot x^4=x^2(3x^2+11)=3x^4+11x^2=3(3x^2+11)+11x^2=20x^2+33,\\ x^8& =x^2\cdot x^6=x^2(20x^2+33)=20x^4+33x^2=20(3x^2+11)+33x^2=93x^2+220,\\ x^{10}& =x^2\cdot x^8=x^2(93x^2+220)=93x^4+220x^2=93(3x^2+11)+220x^2=499x^2+1023.\end{array}$$Thus, $x^{10}=a{x}^6+b{x}^4+c$ becomes $$499x^2+1023=a(20x^2+33)+b(3x^2+11)+c.$$Then $$499x^2+1023=(20a+3b)x^2+(33a+11b+c).$$Since $x^2$ is irrational, we want $a,$ $b,$ and $c$ to satisfy $20a+3b=499$ and $33a+11b+c=1023.$ Solving for $a$ and $b,$ we find $$a=\frac{3c+2420}{121},b=\frac{3993-20c}{121}.$$Hence, $c<\frac{3993}{20},$ which means $c\le 199.$ Also, we want $3c+2420$ to be divisible by 121 Since 2420 is divisible by 121, $c$ must be divisible by 121. Therefore, $c=121,$ which implies $a=23$ and $b=13,$ so $a+b+c=\boxed{157}.$