Solve x2+2x−8>0x^2 + 2x - 8 > 0x2+2x−8>0.A−2<x<4-2 < x < 4−2<x<4Bx<−4x < -4x<−4 or x>2x > 2x>2check_circleC−4<x<2-4 < x < 2−4<x<2Dx<−2x < -2x<−2 or x>4x > 4x>4Explanation(x+4)(x−2)>0(x+4)(x-2) > 0(x+4)(x−2)>0: outside the roots, so x<−4x < -4x<−4 or x>2x > 2x>2.