Question

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Answer 1

${x}^{2} + 6x + 7 = 5 + 4 \sqrt{3} \\ {x}^{2} + 6x + 7 - 5 - 4 \sqrt{3 } = 0 \\ {x}^{2} + 6x + 2 - 4 \sqrt{3} = 0$

$a = 1 \\ b = 6 \\ c = 2 - 4 \sqrt{3}$

$delta = {b}^{2} - 4ac \\ delta = 36 - 4(2 - 4 \sqrt{3)} = 36 - 8 + 16 \sqrt{3 } \\ delta = 28 - 16 \sqrt{3} = 4(7 - 4 \sqrt{3)}$

$x = \frac{ - b + \sqrt{delta} }{2a} = \frac{ - 6 +2 \sqrt{7 - 4 \sqrt{3} } }{2} = - 3 + \sqrt{7 - 4 \sqrt{3} }$

$y = - 3 - \sqrt{7 - 4 \sqrt{3} }$