<s>[INST] Solve $x^2-|5x-3|-x<2,\ \ x\in \mathbb{R} $ 
Solve $x^2-|5x-3|-x<2,\ \ x\in \mathbb{R} $ 

I tried $x^2-|5x-3|-x<2$  ,
case $1$ ,  $x^2-(5x-3)-x<2,\ x\geq 0 \\
x^2-6x+1<0 \\
3-2\sqrt2 < 3+2\sqrt2 \\
 0.17<x<5.8\\
$
$x^2-(5x-3)-x<2$  ,  
case $2$ ,  $x^2+(5x-3)-x<2,\ x< 0 \\
x^2+4x-5<0 \\
-5 < x< 1\\
$
The region common is $3-2\sqrt2<x<1$
But the book gives answer $-5<x<3+2\sqrt2$ .  I am confused.
 [/INST] Case 1 is not for $x>0$ but for $5x-3>0\implies x>\frac{3}{5}$ 
So for case 1 you have $\frac{3}{5}<x<3+\sqrt{2}$ (since $\frac{3}{5}>3-\sqrt{2}$
For case 2 you have $x<\frac{3}{5}$, so $-5<x<\frac{3}{5}$
So the general solution is $-5<x<3+\sqrt{2}$
</s>