Remove square roots by taking squares:
10.
[tex]\sqrt{3x+10}=5-2x[/tex]
[tex](\sqrt{3x+10})^2=(5-2x)^2[/tex]
[tex]3x+10=25-20x+4x^2[/tex]
Keep in mind that [tex]\sqrt x[/tex] is defined as long as [tex]x\ge0[/tex]; in this case, we require [tex]3x+10\ge0[/tex], or [tex]x\ge-\dfrac{10}3[/tex].
[tex]4x^2-23x+15=0[/tex]
[tex](4x-3)(x-5)=0[/tex]
[tex]4x-3=0\text{ or }x-5=0[/tex]
[tex]x=\dfrac34\text{ or }x=5[/tex]
Both of these solutions are greater than -10/3, so they are both valid.
11.
[tex]4x=\sqrt{1-6x}[/tex]
This tells us we need to have [tex]1-6x\ge0[/tex], or [tex]x\le\dfrac16[/tex].
[tex](4x)^2=(\sqrt{1-6x})^2[/tex]
[tex]16x^2=1-6x[/tex]
[tex]16x^2+6x-1=0[/tex]
[tex](8x-1)(2x+1)=0[/tex]
[tex]8x-1=0\text{ or }2x+1=0[/tex]
[tex]x=\dfrac18\text{ or }x=-\dfrac12[/tex]
Both of these are valid solutions.
