2x(x-3) + 15 + 5x = 0
⇒ $2x^{2}-6x+15+5x=0$
⇒ $2x^{2}-x+15=0$
⇒ $2x^{2} - x +$$\frac{1}{(2\sqrt{2})^2}+\frac{119}{8}=0$
⇒ $(\sqrt{2}x+$ $\frac{1}{2\sqrt{2}})^2$ + $\frac{119}{8}$ = 0
Mà $(\sqrt{2}x+$ $\frac{1}{2\sqrt{2}})^2$ ≥ 0 ∀x
⇒ $(\sqrt{2}x+$ $\frac{1}{2\sqrt{2}})^2$ + $\frac{119}{8}$ ≥ $\frac{119}{8}$ ∀x
Vậy pt vô nghiệm.