Answer
$x=\frac{36-12\sqrt 2}{3\sqrt 3+2\sqrt 2-2\sqrt 6}\approx6.09$
Work Step by Step
Step 1. Use the Law of Cosines in triangle CBQ, we have $a^2=6^2+4^2-2\times4\times6cos45^{\circ}=52-24\sqrt 2$
Step 2. Use the Law of Cosines in triangle ACQ, we have $b^2=x^2+6^2-2\times x\times6cos30^{\circ}=x^2+36-6\sqrt 3x$
Step 3. Use the Law of Cosines in triangle ABQ, we have $c^2=x^2+4^2-2\times x\times4cos75^{\circ}=x^2+16-8cos75^{\circ}x$ Here you can use the Addition Formula to get $cos75^{\circ}=\frac{\sqrt 6-\sqrt 2}{2}$, otherwise, just use the numeric value.
Step 4. Use the Pythagorean Theorem in the right triangle of ABC as $c^2=a^2+b^2$, we have
$x^2+16-8cos75^{\circ}x=52-24\sqrt 2+x^2+36-6\sqrt 3x$
Step 5. Solve the above equation to get $x=\frac{36-12\sqrt 2}{3\sqrt 3+2\sqrt 2-2\sqrt 6}\approx6.09$
