Use the distributive property to multiply $-4$ by $10y^{2}-5y-5$.
$$9y^{2}+42y+49-40y^{2}+20y+20\geq 0$$
Combine $9y^{2}$ and $-40y^{2}$ to get $-31y^{2}$.
$$-31y^{2}+42y+49+20y+20\geq 0$$
Combine $42y$ and $20y$ to get $62y$.
$$-31y^{2}+62y+49+20\geq 0$$
Add $49$ and $20$ to get $69$.
$$-31y^{2}+62y+69\geq 0$$
Multiply the inequality by -1 to make the coefficient of the highest power in $-31y^{2}+62y+69$ positive. Since $-1$ is negative, the inequality direction is changed.
$$31y^{2}-62y-69\leq 0$$
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$31y^{2}-62y-69=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $31$ for $a$, $-62$ for $b$, and $-69$ for $c$ in the quadratic formula.
For the product to be $≤0$, one of the values $y-\left(\frac{10\sqrt{31}}{31}+1\right)$ and $y-\left(-\frac{10\sqrt{31}}{31}+1\right)$ has to be $≥0$ and the other has to be $≤0$. Consider the case when $y-\left(\frac{10\sqrt{31}}{31}+1\right)\geq 0$ and $y-\left(-\frac{10\sqrt{31}}{31}+1\right)\leq 0$.
