\[\boxed{\text{1078\ (1078).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[x² + 2\sqrt{3x} + y - 4\sqrt{y} + 7 = 0\]
\[D = 12 - 4 \cdot \left( y - 4\sqrt{y} + 7 \right) =\]
\[= 12 - 4y + 16\sqrt{y} - 28 =\]
\[= - 4y + 16\sqrt{y} - 16 =\]
\[= - 4 \cdot \left( \sqrt{y} - 2 \right)^{2}.\]
\[D \geq 0,\ \ если\ \ \ \left( \sqrt{y} - 2 \right)^{2} = 0,\]
\[\sqrt{y} = 2.\]
\[x^{2} + 2\sqrt{3x} + 4 - 8 + 7 = 0\]
\[x^{2} + 2\sqrt{3x} + 3 = 0\]
\[x = - \frac{2\sqrt{3}}{2} = - \sqrt{3},\ \ y = 4.\]
\[Ответ:x = - \sqrt{3};\ \ \ y = 4.\]