Consider the following quadratic equations in $x$ $$x^2+2x-15=0 \tag{1}$$ $$2x^2+3x+a^2+12a=0 \tag{2}$$ Let us denote the two solutions of (1) by $\alpha$ and $\beta$ ($\alpha < \beta$). We are to find the range of values which $a$ in (2) can take, in order that (2) has two real solutions $\gamma$ and $\delta$ and they satisfy $$\alpha < \gamma < \beta < \delta.$$ (1) $\alpha = \mathbf{AB}$ and $\beta = \mathbf{C}$. (2) When we set $b = a^2+12a$, from the condition $\alpha < \gamma$ we have $$b > \mathbf{DEF}$$ and from the condition $\gamma < \beta < \delta$ we have $$b < \mathbf{GHI}.$$ Hence the range of the values which $a$ can take is $$\mathbf{JK} < a < \mathbf{LM}, \quad \mathbf{NO} < a < \mathbf{PQ},$$ where $\mathrm{JK} < \mathrm{NO}$.
Consider the following quadratic equations in $x$
$$x^2+2x-15=0 \tag{1}$$
$$2x^2+3x+a^2+12a=0 \tag{2}$$
Let us denote the two solutions of (1) by $\alpha$ and $\beta$ ($\alpha < \beta$). We are to find the range of values which $a$ in (2) can take, in order that (2) has two real solutions $\gamma$ and $\delta$ and they satisfy
$$\alpha < \gamma < \beta < \delta.$$
(1) $\alpha = \mathbf{AB}$ and $\beta = \mathbf{C}$.
(2) When we set $b = a^2+12a$, from the condition $\alpha < \gamma$ we have
$$b > \mathbf{DEF}$$
and from the condition $\gamma < \beta < \delta$ we have
$$b < \mathbf{GHI}.$$
Hence the range of the values which $a$ can take is
$$\mathbf{JK} < a < \mathbf{LM}, \quad \mathbf{NO} < a < \mathbf{PQ},$$
where $\mathrm{JK} < \mathrm{NO}$.