3х^2-4х-4=0
a=3
b=-4
c=-4
calculam Δ
Δ=b^2-4ac
Δ=(-4)^2-4*3*(-4)
Δ=16+48
Δ=64=8^2
x₁=(-b-√Δ)/2a=(4-8)/6=-4/6=-2/3
x₂=(-b+√Δ)/2a=(4+8)/6=12/6=2
[tex]\it 3x^2-4x-4=0 \Rightarrow 3x^2+2x-6x-4=0 \Rightarrow x(3x+2)-2(3x+2)=0 \Rightarrow \\ \\ \Rightarrow (3x+2)(x-2)=0 \Rightarrow \begin{cases}\it 3x+2=0 \Rightarrow x_1=-\dfrac{2}{3}\\ \\ \it x-2=0 \Rightarrow x_2=2\end{cases}[/tex]
