Geometric sequence problem including sum of the numbers Numbers: $a,b,c,d$ generate geometric sequence and $a+b+c+d=-40. $ Find these numbers if $a^2+b^2+c^2+d^2=3280$
I tried this problem and I have system of equations which I can't solve. I think there should be different way to handle this problem.

$$-48=\dfrac p{d^3}+\dfrac pd+pd+pd^3=\dfrac p{d^3}(1+d^2+d^4+d^6)=\dfrac{p(1+d^2)(1+d^4)}{d^3}$$
Similarly,
$$3280=\dfrac{p^2(1+d^4)(1+d^8)}{d^6}$$
$$\implies\dfrac{(-48)^2}{3280}=\dfrac{(1+d^2)^2(1+d^4)}{1+d^8}=\dfrac{\left(d+\dfrac1d\right)^2\left(d^2+\dfrac1{d^2}\right)}{d^4+\dfrac1{d^4}}$$
Let $d^2+\dfrac1{d^2}=u$
$$d^4+\dfrac1{d^4}=u^2-2, \left(d+\dfrac1d\right)^2=u+2$$
So, we have a quadratic equation in $u$
