Answer: \(x=-3,\frac52,\frac{-1+\sqrt{41}}4,\frac{-1-\sqrt{41}}4\).

The equation

\(\left|2x^2+x-10\right|=5\)

splits into two equations:

\(2x^2+x-10=5\quad\text{or}\quad 2x^2+x-10=-5.\)

First solve

\(2x^2+x-15=0.\)

Using the quadratic formula,

\(x=\dfrac{-1\pm\sqrt{1+120}}{4}=\dfrac{-1\pm11}{4}.\)

So

\(x=\dfrac52\quad\text{or}\quad x=-3.\)

Now solve

\(2x^2+x-5=0.\)

Again using the quadratic formula,

\(x=\dfrac{-1\pm\sqrt{1+40}}{4}=\dfrac{-1\pm\sqrt{41}}{4}.\)

Therefore the complete set of solutions is

\(x=-3,\quad \dfrac52,\quad \dfrac{-1+\sqrt{41}}4,\quad \dfrac{-1-\sqrt{41}}4.\)

Answer: \(x=-1,2,3,6\).

The equation

\(\left|x^2-5x\right|=6\)

means that the expression inside the modulus is either \(6\) or \(-6\).

First solve

\(x^2-5x=6.\)

This gives

\(x^2-5x-6=0,\)

so

\((x-6)(x+1)=0.\)

Hence

\(x=6\quad\text{or}\quad x=-1.\)

Now solve

\(x^2-5x=-6.\)

This gives

\(x^2-5x+6=0,\)

so

\((x-2)(x-3)=0.\)

Hence

\(x=2\quad\text{or}\quad x=3.\)

Therefore the complete set of solutions is

\(x=-1,\ 2,\ 3,\ 6.\)