There is the following equation.
$$x^{2}+2(m-a)x+3am-2=0$$
a) Find $a$ such that the equation has real roots, $\forall m\in \mathbb{R}$
b) Find $m$ such that the equation has real roots, $\forall a\in \mathbb{R}$
The discriminant is $m^{2}-5am+a^{2}+2\geq 0$ and $a^{2}-5am+m^{2}+2\geq 0$
I found that $a,m\in (-\infty , -\sqrt{\frac{8}{21}}]\cup [\sqrt{\frac{8}{21}}, +\infty)$ but I don't know why $|a|\leqslant \sqrt{\frac{8}{21}}$and $|m|\leqslant \sqrt{\frac{8}{21}}$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in $a$ is $\le0$. For radicals I know that it should be $\ge0$.
So you have a discriminant $$\Delta_1(m,a)=m^2-5am+a^2+2$$ and you want it to be greater than zero for all $m$ values. This expression in $m$ with $a$ as parameter is a quadratic. If it is always positive, it means that for $m=0$ you have $\Delta_1(0,a)>0$ (which is true). But you want to be true for all $m$ values. If this expression would have real roots, there are some values where the discriminant is negative, so the original equation does not have real solutions. So $$\Delta_2(a)=25 a^2-4(a^2+2)<0$$ This will ensure that $\Delta_1(m,a)>0$ for any $m$.