1. **State the problem:** We want to find a whole number $x$ such that the sum of $x$ and twice the square of $x$ equals 10. 2. **Write an equation:** The problem translates to the equation $$x + 2x^2 = 10.$$ 3. **Rearrange the equation:** Move all terms to one side to set the equation to zero: $$2x^2 + x - 10 = 0.$$ 4. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where $a=2$, $b=1$, and $c=-10$. 5. **Calculate the discriminant:** $$b^2 - 4ac = 1^2 - 4 \times 2 \times (-10) = 1 + 80 = 81.$$ 6. **Find roots:** $$x = \frac{-1 \pm \sqrt{81}}{2 \times 2} = \frac{-1 \pm 9}{4}.$$ 7. **Evaluate each root:** - First root: $$x = \frac{-1 + 9}{4} = \frac{8}{4} = 2.$$ - Second root: $$x = \frac{-1 - 9}{4} = \frac{-10}{4} = -2.5.$$ 8. **Determine whole number:** Since whole numbers are non-negative integers (0,1,2,...), only $$x=2$$ is valid. **Final answer:** The number is $2$.