$4x^2-6mx+\left(2m+3\right)\left(m-3\right)=0$  

Równanie ma dwa różne rozwiązania, gdy: 

$\Delta >0$  

${\left(-6m\right)}^2-4\cdot 4\cdot \left(2m+3\right)\left(m-3\right)>0$  

$36m^2-16\left(2m^2-6m+3m-9\right)>0\mid :4$  

$9m^2-4\left(2m^2-3m-9\right)>0$  

$9m^2-8m^2+12m+36>0$  

$m^2+12m+36>0$  

${\left(m+6\right)}^2>0$  

$m\in \mathbb{R}-\left\{-6\right\}$  

 

Spełniony jest warunek:

$\left(4x_1-4x_2-1\right)\left(4x_1-4x_2+1\right)<0$  

${\left(4x_1-4x_2\right)}^2-1^2<0$  

${\left(4\left(x_1-x_2\right)\right)}^2<1$  

$16\left(x_1^2-2x_1{x}_2+x_2^2\right)<1$  

$x_1^2-2x_1{x}_2+x_2^2<\frac{1}{16}$  

${\left(x_1+x_2\right)}^2-2x_1{x}_2-2x_1{x}_2<\frac{1}{16}$  

${\left(x_1+x_2\right)}^2-4x_1{x}_2<\frac{1}{16}$  

${\left(\frac{-b}{a}\right)}^2-4\cdot \frac{c}{a}<\frac{1}{16}$  

$\frac{b^2}{a^2}-4\cdot \frac{c}{a}<\frac{1}{16}$  

$\frac{{\left(-6m\right)}^2}{16}-4\cdot \frac{\left(2m+3\right)\left(m-3\right)}{4}<\frac{1}{16}\mid \cdot 16$  

$36m^2-16\left(2m+3\right)\left(m-3\right)<1$  

$36m^2-16\left(2m^2-6m+3m-9\right)<1$  

$36m^2-16\left(2m^2-3m-9\right)<1$  

$36m^2-32m^2+48m+144<1$  

$4m^2+48m+143<0$  

${\Delta }_m={48}^2-4\cdot 4\cdot 143=2304-2288=16$  

$m_1=\frac{-48-4}{2\cdot 4}=\frac{-52}{8}=-\frac{13}{2}$  

$m_2=\frac{-48+4}{2\cdot 4}=\frac{-44}{8}=-\frac{11}{2}$  

$m\in \left(-\frac{13}{2},-\frac{11}{2}\right)$  

Uwzględniając wcześniejsze założenie otrzymujemy:

$m\in \left(-\frac{13}{2},-6\right)\cup \left(-6,-\frac{11}{2}\right)$