Answer
$x=0, 4$
Work Step by Step
$\sqrt {3x+4} -1=\sqrt {2x+1}$
$\sqrt {3x+4} -1+1=\sqrt {2x+1}+1$
$\sqrt {3x+4} =\sqrt {2x+1}+1$
$(\sqrt {3x+4})^2 =(\sqrt {2x+1}+1)^2$
$3x+4=(\sqrt {2x+1})(\sqrt {2x+1})+1*(\sqrt {2x+1})+1*(\sqrt {2x+1})+1*1$
$3x+4=2x+1+2\sqrt {2x+1}+1$
$3x+4=2x+2+2\sqrt {2x+1}$
$x+2=2\sqrt {2x+1}$
$(x+2)^2=(2\sqrt {2x+1})^2$
$x*x+2*x+x*2+2*2=2*2*(\sqrt{2x+1})^2$
$x^2+4x+4=4(2x+1)$
$x^2+4x+4=8x+4$
$x^2-4x=0$
$x(x-4)=0$
$x=0$
$x-4=0$
$x=4$
$\sqrt {3x+4} -1=\sqrt {2x+1}$
$\sqrt {3*0+4} -1=\sqrt {2*0+1}$
$\sqrt {4} -1=\sqrt {1}$
$2-1=1$ (true)
$\sqrt {3x+4} -1=\sqrt {2x+1}$
$\sqrt {3*4+4} -1=\sqrt {2*4+1}$
$\sqrt {12+4} -1=\sqrt {8+1}$
$\sqrt {16} -1=\sqrt {9}$
$4-1=3$ (true)
