Решите неравенство 1/(2x+3)^2≥4

$\frac{1}{(2x+3)^2}\geq4|*(2x+3)^2\\ x\neq -\frac{3}{2}; x\neq-1.5\\1\geq 4(2x+3)^2\\4(2x+3)^2\leq 1\\4(4x^2+12x+9)\leq 1\\16x^2+48x+36\leq 1\\16x^2+48x+36-1\leq 0\\16x^2+48x+35\leq 0\\D^2=48^2-4*16*35=2304-2240=64; D=\sqrt{64}=+-8\\ x_1=\frac{-48+8}{2*16}=\frac{-40}{32}=\frac{-10}{8}=-\frac{5}{4}=-1.25\\\\x_2=\frac{-48-8}{2*16}=\frac{-56}{32}=-\frac{7}{4}=-1.75\\16(x+1.25)(x+1.75)\leq 0\\$

(+) (-) (+)

____-1.75______-1.25___0

xє[ -1.75;-1.25]