Question

help plz...... help me

Answer 1

$\sqrt{3} {x}^{2} - 2 \sqrt{2x} - 2 \sqrt{3} x = 0 \\ a {x}^{2} + bx + c \\ a = \sqrt{3} \: \: \: \: \: \: \: \: \: b = - 2 \sqrt{2} \: \: \: \: \: \: \: c = - 2 \sqrt{3} \\ \frac{ - b + or - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \frac{2 \sqrt{2} + or - \sqrt{ {(2 \sqrt{2}) }^{2} - 4( \sqrt{3})( - 2 \sqrt{3} )} }{2 \sqrt{3} } \\ \frac{2 \sqrt{2} + or - \sqrt{8 + 24} }{2 \sqrt{3} } \\ \frac{2 \sqrt{2} + or - \sqrt{32} }{2 \sqrt{3} } \\ \frac{2 \sqrt{2} + or - 4\sqrt{2} }{2 \sqrt{3} } \\ \frac{2 \sqrt{2} + 4 \sqrt{2} }{2 \sqrt{3} } \\ \frac{6 \sqrt{2} }{2 \sqrt{3} } = \frac{3 \sqrt{2} }{ \sqrt{3} } \\ second \: answer \\ \frac{2 \sqrt{2} - 4 \sqrt{2} }{2 \sqrt{3} } \\ - \frac{2 \sqrt{2} }{2 \sqrt{3} } = \frac{\sqrt{2} }{ \sqrt{3} }$