Risolvi (9+4m^2)x^2+(m^2-m)32x+(64m^2+38)=0

$$( 9 + 4 m ^ { 2 } ) x ^ { 2 } + ( m ^ { 2 } - m ) 32 x + ( 64 m ^ { 2 } + 38 ) = 0$$

Trova m

$$\left\{\begin{matrix}m=\frac{\sqrt{-9x^{4}-72x^{3}-118x^{2}-304x-608}+8x}{2\left(x+4\right)^{2}}\text{; }m=\frac{-\sqrt{-9x^{4}-72x^{3}-118x^{2}-304x-608}+8x}{2\left(x+4\right)^{2}}\text{, }&x\neq -4\text{ and }-9x^{4}-72x^{3}-118x^{2}-304x-608\geq 0\\m=-\frac{91}{64}=-1,421875\text{, }&x=-4\end{matrix}\right,$$

Trova x

$$x=\frac{-16m^{2}+\sqrt{2\left(-256m^{3}-236m^{2}-171\right)}+16m}{4m^{2}+9}$$

$$x=\frac{-16m^{2}-\sqrt{2\left(-256m^{3}-236m^{2}-171\right)}+16m}{4m^{2}+9}\text{, }-2048m^{3}-1888m^{2}-1368\geq 0$$