Решите неравенство методом интервалов:* < 6

$\sqrt{x+3} * \sqrt{4x+5} \ \textless \ 6$ ; _________________________ $\sqrt{(x+3)(4x+5)} \ \textless \ \sqrt{36}$ ; _______________________ $(x+3)(4x+5) \ \textless \ 36$ ; ________________________________ $4x^2+5x+12x+15-36\ \textless \ 0$ ; ____________________________ $4x^2+17x-21\ \textless \ 0$ ; _________________________________ $4x^2+17x-21=0; D=17^2-4*4*(-21)=289+336=625=25^2$ ; ____________ $x_1= \frac{-17-25}{2*4} =-5,25; x_2= \frac{-17+25}{2*4} =1$ ; ___________ _______ + ____________________ - ____________________ + ________ -----------------------------(-5,25)----------------------------------------(1)--------------------------------- x∈(-5,25;1)