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20-01-2021
Mathematics

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[tex](x +3)^{2} + 10x \leqslant 21[/tex]

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asmaNasr asmaNasr

20-01-2021

Step-by-step explanation:

[tex] {(x + 3)}^{2} + 10x\leqslant 10 \\ {x}^{2} + 6x + 9 + 10x \leqslant 10 \\ {x}^{2} + 16x + 9 \leqslant 10 \\ {x}^{2} + 6x + 9 \leqslant 0 \\ the \: value \: of \: x \: that \: makes \: it \: exist \\ x = - 3[/tex]

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