Question

help me pls pls pls pls pls pls pls pls pls pls pls ​

Answer 1

Ответ:

$\sqrt{3 {x }^{2} + 3x - 9 } = 1 \\ 3 {x}^{2} + 3x - 9 = 1 \\ 3 {x}^{2} + 3x - 10 = 0 \\ d = 9 + 120 = 129 \\ x1 = \frac{ - 3 + \sqrt{129} }{6} \\ x2 = \frac{ - 3 - \sqrt{129} }{6}$