Пусть x и y – корни уравнения 3x^2 -5x -3 = 0. x^4+y^4=?

Решите задачу:

$3x^2-5x-3=0\\\\D=25+4\cdot 3\cdot 3=61\\\\x_1= \frac{5-\sqrt{61}}{6} \; ,\; \; x_2= \frac{5+\sqrt{61}}{6} \\\\x_1^4+x_2^4=x^4+y^4= \frac{(5-\sqrt{61})^4}{6^4}+ \frac{(5+\sqrt{61})^4}{6^4} = \frac{((5-\sqrt{61})^2)^2+((5+\sqrt{61})^2)^2}{y} =\\\\= \frac{(86-10\sqrt{61})^2+(86+10\sqrt{61})^2}{1296} = \frac{86^2-1720\sqrt{61}+61+86^2+1720\sqrt{61}+61}{1296} =\\\\= \frac{14914}{1296} =11 \frac{329}{648}$