Помогите решить уравнение 10^(1+x^2)-10^(1-x^2)=99

Решите задачу:

$10^{1+x^2}-10^{1-x^2}=99 \\Range: x \in R \\10*10^{x^2}-\frac{10}{10^{x^2}}=99 \\\frac{10*10^{2x^2}-10}{10^{x^2}}=99 \\99*10^{x^2}=10*10^{2x^2}-10 \\10*(10^{x^2})^2-99*10^{x^2}-10=0 \\Substitution: 10^{x^2}=t,\ range: t\ \textgreater \ 0 \\10t^2-99t-10=0 \\D=b^2-4ac=9801-4*10*(-10)=9801+400=10201 \\t=\frac{99+101}{20}\ \ or\ \ t=\frac{99-101}{20} \\t=10\ \ or\ \ t=-\frac{1}{10}\ (out\ of\ range) \\Reverse\ substitution: 10^{x^2}=10 \\x^2=1 \\Answer: x=1\ or\ x=-1$