Answer
$x=-\displaystyle \frac{2}{9}$ or $x=2$
Work Step by Step
We are given:
$\sqrt{2\sqrt{7x+2}}=\sqrt{3x+2}$
We square both sides:
$(\sqrt{2\sqrt{7x+2}})^{2}=(\sqrt{3x+2})^{2}$
$2\sqrt{7x+2}=3x+2$
And square both sides again:
$(2\sqrt{7x+2})^{2}=(3x+2)^{2}$
$(2\sqrt{7x+2})(2\sqrt{7x+2})=(3x+2)(3x+2)$
$4\ (7x+2)=9x^{2}+12x+4$
And distribute:
$28x+8=9x^{2}+12x+4$
$9x^{2}-16x-4=0$
And factor:
$(9x+2)(x-2)=0$
Use the zero-factor property by equating each factor to zero:
$(9x+2)=0$ or $(x-2)=0$
$x=-\displaystyle \frac{2}{9}$ or $x=2$