Quadratic formula: where ax^2+bx+c =0, x = (-b±√(b^2-4ac))/2a.Rearrange the formula in the question to get 3x^2+4x-20 = 0.From the quadratic in the question we can see that a = 3, b = 4, c = -20.Therefore x = (-4±√(4^2 - (43-20)))/(2*3)= (-4±√(16 - -240))/6 = (-4±√256)/6= (-4±16)/6= -20/6 AND 12/6= -10/3 AND 2Therefore x = -3.3r AND 2.