Hello,
[tex] \sqrt{6-5x} +3=4x\\\\
\sqrt{6-5x} =4x-3 \\\\
6-5x=(4x-3)^2\\\\
6-5x=16x^2-24x+9\\\\
16x^2-19x+3=0\\\\
\Delta=19^2-4*3*19=169=13^2\\\\
x= \dfrac{19+13}{32} =1\ or\ x=\dfrac{19-13}{32} = \dfrac{3}{16} \\\\
If\ x=1\ then\ \sqrt{6-5} +3=4*1\ ==\ \textgreater \ 1+3=4 \\\\
If\ x=\dfrac{3}{16}\ then\ \sqrt{6-5*\dfrac{3}{16}} +3=4*\dfrac{3}{16}\ ==\ \textgreater \ \dfrac{21}{4} \neq \dfrac{3}{4} \\
Solution=\{1\}
[/tex]
