x в квадрате + 4х - 12 > 5x
5x\\x^2-x-12>0\\x_1=4\ ;x_2=-3" alt="x^2 + 4x - 12 > 5x\\x^2-x-12>0\\x_1=4\ ;x_2=-3" align="absmiddle" class="latex-formula"> ///+///(-3)...-...(4)///+///=>x
x ² + 4х - 12 > 5x
x ² + 4х - 12 - 5x>,0
x ² -х - 12 > 0
x ² -х - 12 = 0
D = 1+48 =49
x₁= 4 , x₂ = -3
Рисунок во вложении
Ответ : х ∈ ( -∞ ; -3) υ (4; +∞)
5x\\x^2-x-12>0\\x_1=4\ ;x_2=-3" alt="x^2 + 4x - 12 > 5x\\x^2-x-12>0\\x_1=4\ ;x_2=-3" align="absmiddle" class="latex-formula">
