Арифметическое прогрессия а1+а2+а3=27 а1^2+а2^2+а3^2=275 а1,а2,а3-?

Решите задачу:

$a_{1}+a_{1}+d+a_{1}+2d=27\\ a_{1}^2+(a_{1}+d)^2+(a_{1}+2d)^2=275\\\\ 3a_{1}+3d=27\\\ a_{1}+d=9\\ a_{2}=9\\\\ a_{1}^2+(a_{1}+2*(9-a_{1}))^2=194\\ a_{1}^2+(18-a_{1})^2=194\\ 2a_{1}^2+324-36a_{1}=194\\ 2a_{1}^2-36a_{1}+130=0\\ a_{1}^2-18a_{1}+65=0\\ D=324-4*1*65=8^2\\ a_{1}=\frac{18+8}{2}=13\\ a_{1}=\frac{18-8}{2}=5\\ a_{1}=13; \ \ \ d=-4\\ a_{1}=5; \ \ \ d=4\\\\ a_{2}=9\\ a_{3}=5\\ a_{2}=9\\ a_{3}=13$