Решите уравнения ; 0,6^2x^2+8x<1 нужна помощь? плииз

Решите задачу:

$0,6^2x^2+8x\ \textless \ 1\\\\0,36x^2+8x-1\ \textless \ 0\\\\D/4=16+0,36=16,36=\frac{1636}{100} \\\\\sqrt{D/4}= \frac{\sqrt{1636}}{10}=\frac{2\sqrt{409}}{10} =\frac{\sqrt{409}}{5}\\\\x_1= \frac{-4-\sqrt{409}/5}{0,36} =\frac{-20-\sqrt{409}}{1,8}=\frac{-20-\sqrt{409}}{9/5}= \frac{-100-5\sqrt{409}}{9} \\\\x_2= \frac{-100+5\sqrt{409}}{9} \\\\x\in \Big (\frac{-100-5\sqrt{409}}{9}\; ;\; \frac{-100+5\sqrt{409}}{9}\Big )$