Part 1
we have the equation
[tex]\mleft|2x+1\mright|-2=3x+2[/tex]Solve for x
[tex]\begin{gathered} |2x+1|=3x+2+2 \\ |2x+1|=3x+4 \end{gathered}[/tex]REmember that the absolute value function has two solutions
First solution (case positive)
[tex]\begin{gathered} +(2x+1)=3x+4 \\ 3x-2x=1-4 \\ x=-3 \end{gathered}[/tex]Second solution (negative case)
[tex]\begin{gathered} -(2x+1)=3x+4 \\ -2x-1=3x+4 \\ 3x+2x=-1-4 \\ 5x=-5 \\ x=-1 \end{gathered}[/tex]therefore
the solutions are x=-3 and x=-1
Part 2
we have the function
[tex]f(x)=-\mleft|x^3-2x^2\mright|-2[/tex]Remember that
f(-2) is the value of f(x) when the value of x=-2
substitute the value of x in the expression above
[tex]\begin{gathered} f(-2)=-|-2^3-2(-2)^2|-2 \\ f(-2)=-|-8-8|-2 \\ f(-2)=-|-16|-2 \\ f(-2)=-(16)-2 \\ f(-2)=-18 \end{gathered}[/tex]