[tex]2(x-3)(x+5)>0\\2x^2+4x-30>0\\\Delta=4^2-4\cdot2\cdot(-30)\\\Delta=16+240\\\Delta=256\\\sqrt{\Delta} =\sqrt{256} =16\\x_1=\frac{-4-16}{2\cdot2} =\frac{-20}{4} =-5\\x_2=\frac{-4+16}{2\cdot2} =\frac{12}{4} =3\\x\in(-\infty,-5)\cup(3,+\infty)[/tex]
