Найти область определения функции

$y = \sqrt{x+1} + \sqrt[6]{x^{2}+2 x-3 }$

$\left \{ {{x+1 \geq 0} \atop {x^2+2x-3 \geq 0}} \right.$

$x^2+2x-3=0$
$D=2^2-4*1*(-3)=4+12=16$
$x_1= \frac{-2+4}{2}=1$
$x_2= \frac{-2-4}{2}=-3$
$\left \{ {{x+1 \geq 0} \atop {(x-1)(x+3) \geq 0}} \right.$
$\left \{ {{x \geq -1} \atop {(x-1)(x+3) \geq 0}} \right.$

+ - +
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Ответ: $D(y)=$ $[1;+$ ∞ $)$