Respuesta :
[tex] - 4 \sqrt{3} x - 2 + 6 = 22 \\ - 4 \sqrt{3}x + 4 = 22 \\ - 4 \sqrt{3}x = 18 \\ x = \frac{18}{ - 4 \sqrt{3} } = \frac{ - 18 \sqrt{3} }{12} = \frac{ - 3 \sqrt{3} }{2} = - 1.5 \sqrt{3} [/tex]
HOPE THIS WILL HELP YOU
The solution depends on the argument of the square root. Please be more precise and less ambiguous when writing your questions.
You could either mean:
[tex]-4\sqrt{3}x-2+6=22,\quad -4\sqrt{3x}-2+6=22,\quad -4\sqrt{3x-2}+6=22[/tex]
In the first case, we have
[tex]-4\sqrt{3}x-2+6=22 \iff -4\sqrt{3}x= 18 \iff x = -\dfrac{18}{4\sqrt{3}}[/tex]
In the second case, we have
[tex]-4\sqrt{3x}-2+6=22 \iff \sqrt{3x}=-\dfrac{9}{2}[/tex]
which has no solution, because a square root can't be negative
In the third case, we have
[tex]-4\sqrt{3x-2}+6=22 \iff -4\sqrt{3x-2}=16 \iff \sqrt{3x-2}=-4[/tex]
which again has no solution, for the same reason.