In Q14 of section 2.6, we are given the initial value problem and solve for at the approximate period where the solution is valid. The equation is

$$(9x^2+y−1)−(4y−x)y′=0,$$

with initial value $y(1)=0$. I have solved the equation that

$$3x^3+yx-x-2y^2=2,$$

and this is in line with the answer. But I don't understand why we take the quadratic formula in term of $y$ to find the solution range and even if we do, how can we solve

$$24x^3+x^2-8x-16 \ge 0$$

without the help of a calculator? Thanks for your help.