Answer
Solution set:$ \; $ $\displaystyle \{\frac{1}{2},1\}.$
Work Step by Step
$10x-1=(2x+1)^{2}$
$10x-1=4x^{2}+4x+1$
$0=4x^{2}-6x+2$
$0=2x^{2}-3x+1$
Factor by searching for factors of $ ac=2$ whose sum is $ b=-3$; we find $-1$ and $-2$. Rewrite $ bx $ and factor in pairs:
$0=2x^{2}-2x-x+1$
$0=2x(x-1)-(x-1)$
$0=(x-1)(2x-1)$
$ x=\displaystyle \frac{1}{2}, \; x=1$
Solution set:$ \; $ $\displaystyle \{\frac{1}{2},1\}.$
