Question

${x}^{2} + 2x - 48 < 0$
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Answer 1

Смотри....................,

Answer 2

Решение равенства

$x^{2} +2x-48$$\leq 0$

$x^{2} +8x-6x-48\leq 0$

$x(x+8)-6(x+8)\leq 0$

$(x+8)(x-6)\leq 0$

$\left \{ {{x+8\leq 0} \atop {x-6\geq 0}} \right.$

$\left \{ {{x+8\geq 0} \atop {x-6\leq 0}} \right.$

$\left \{ {{x\leq -8} \atop {x\geq 6}} \right.$

$\left \{ {{x\geq -8} \atop {x\leq 6}} \right.$

$x$∈∅

$x$∈$(-8 , 6)$