Дано:S2=2;S6=42.Найти знаменатель геометрической прогрессии-q.Помогите пожалуйста

$S_2=2$
$S_6=42$
$q-$ ?

$S_n= \frac{b_1*(1-q^n)}{1-q} ,$ $q \neq 1$

$\left \{ {{S_2= \frac{b_1*(1-q^2)}{1-q} } \atop {{{S_6= \frac{b_1*(1-q^6)}{1-q} } \right.$
$\left \{ {{ \frac{b_1*(1-q^2)}{1-q} =2} \atop {{{ \frac{b_1*(1-q^6)}{1-q}=42 } \right.$
$\left \{ {{ \frac{b_1*(1-q)(1+q)}{1-q} =2} \atop {{{ \frac{b_1*(1-q^3)(1+q^3)}{1-q}=42 } \right.$
$\left \{ {{b_1*(1+q)} =2} \atop {{{ \frac{b_1*(1-q)(1+q+q^2)(1+q^3)}{1-q}=42 } \right.$
$\left \{ {{b_1*(1+q)} =2} \atop {{b_1*(1+q+q^2)(1+q^3)}=42 } \right.$
$\left \{ {{b_1*(1+q)} =2} \atop {{b_1*(1+q)(1-q+q^2)(1+q+q^2)}=42 } \right.$
$\left \{ {{b_1*(1+q)} =2} \atop {{2*(1-q+q^2)(1+q+q^2)}=42 } \right.$
$\left \{ {{b_1*(1+q)} =2} \atop {{2*((1+q^2)^2-q^2)}=42 } \right.$
$\left \{ {{b_1*(1+q)} =2} \atop {{((1+q^2)^2-q^2)}=21 } \right.$
$\left \{ {{b_1*(1+q)} =2} \atop {{1+q^2+2q-q^2}=21 } \right.$
$\left \{ {{b_1*(1+q)} =2} \atop {{1+2q}=21 } \right.$
$1+2q}=21$
$2q}=21-1$
$2q=20$
$q=10$