$\displaystyle \left (\frac{\pi-3 }{2} \right )^{x^2-x}+\frac{6\pi}{4}-\frac{\pi^{2}+9}{4}> 0$
What I could do
Probably meant$\displaystyle \displaystyle\left (\dfrac{\pi-3 }{2} \right )^{x^2-x}>\dfrac{6\pi}{4}-\dfrac{\pi^{2}+9}{4}$
Corrected in the next line, though.$\displaystyle \displaystyle\left (\dfrac{\pi-3 }{2} \right )^{x^2-x}>\dfrac{\pi^{2}+9}{4}-\dfrac{6\pi}{4}$