[tex]x^{2} -10x+q=0\\D = 100 - 4q\\x_1 = \frac{10+\sqrt{100-4q} }{2}\\ x_2 = \frac{10-\sqrt{100-4q} }{2}\\x_2 - x_1 = 14\\\frac{10-\sqrt{100-4q} }{2} - \frac{10+\sqrt{100-4q} }{2} = 14\\\frac{10-\sqrt{100-4q} -(10+\sqrt{100-4q}) }{2}=14\\10-\sqrt{100-4q} -10-\sqrt{100-4q} = 7\\-2\sqrt{100-4q} = 7\\\sqrt{100-4q} = -14\\100-4q =196\\4q = 100-196\\4q = -96\\q = -24\\x_1 = \frac{10+\sqrt{100-4*(-24)} }{2} = \frac{10+\sqrt{196} }{2} = \frac{10+14 }{2} = \frac{24}{2} = 12\\[/tex]

[tex]x_2 = \frac{10-\sqrt{100-4*(-24)} }{2} = \frac{10-\sqrt{196} }{2} = \frac{10-14}{2} = \frac{-4}{2} = -2[/tex]

Ответ: q = −24; x₁ = 12; x₂ = −2.