Răspuns:
[tex] {x}^{2} + 4 \times x + 4 = 7 + 4 \sqrt{3} \\ ( {x + 2)}^{2} = 7 + 4 \sqrt{3} \\ 7 + 4 \sqrt{3} = 7 + 2 \times 2 \sqrt{3} = 4 + 3 + 2 \times 2 \sqrt{3} = {2}^{2} + 2 \times 2 \times \sqrt{3} + ( { \sqrt{3} )}^{2} = ( {2 + \sqrt{3}) }^{2} \\ = >{(x + 2)}^{2} = {( \sqrt{3} + 2)}^{2} \\ = > x + 2 = \sqrt{3} + 2 = > x = \sqrt{3} \\ sau \: \: x + 2 = - \sqrt{3} - 2 = > x = - \sqrt{3} - 4 [/tex]
