Помогите решить заранее спасибо!

Решите задачу:

$\left \{ {{b_5+b_3=60} \atop {b_7-b_5=144}} \right. \; \left \{ {{b_1q^4+b_1q^2=60} \atop {b_1q^6-b_1q^4=144}} \right. \; \left \{ {{b_1q^2(q^2+1)=60} \atop {b_1q^4(q^2-1)=144}} \right. \\\\ \frac{b_1q^4(q^2-1)}{b_1q^2(q^2+1)}= \frac{144}{60}\\\\q^2(q^2-1)=2,4(q^2+1)\\\\q^4-q^2=2,4q^2+2,4\\\\q^4-3,4q^2-2,4=0\; |\cdot 5\\\\5q^4-17q^2-12=0\; \; \to \; \; t=q^2\; ,\; \; 5t^2-17t-12=0\\\\D=529\; ,\; \; t_1= \frac{17-23}{10}=- \frac{3}{5}\; ,\; t_2=\frac{17+23}{10}=4\\\\q^2=-\frac{3}{5}\ \textless \ 0\; \; net\; kornej$

$q^2=4\; \; \; \to \; \; \; q=\pm 2\\\\ b_1q^2(q^2+1)=60\; \; \to \; \; b_1\cdot 4\cdot 5=60\; ,\; b_1=3\\\\b_2=b_1q=\pm 3\cdot 2=\pm 6\\\\b_4=b_1q^3=\pm 3\cdot 8=\pm 24\\\\b_2\cdot b_4=(\pm 6)\cdot (\pm 24)=+144$