Answer :
From the text of the task we can write the equation:
[tex]3 x^{2} =2(x+4)\\ \\ 3x^2=2x+8\\ \\ 3x^2-2x-8=0\\ \\ \Delta=(-2)^2-4.3.(-8)=4+96=100\\ \\ x=\frac{2 \pm\sqrt{100}}{2.3}=\frac{2 \pm10}{6}\\ \\ x_1=\frac{2-10}{6}=-\frac{8}{6}=-\frac{4}{3}\\ \\ x_2=\frac{2+10}{2.3}=\frac{12}{6}=2[/tex]
[tex]3 x^{2} =2(x+4)\\ \\ 3x^2=2x+8\\ \\ 3x^2-2x-8=0\\ \\ \Delta=(-2)^2-4.3.(-8)=4+96=100\\ \\ x=\frac{2 \pm\sqrt{100}}{2.3}=\frac{2 \pm10}{6}\\ \\ x_1=\frac{2-10}{6}=-\frac{8}{6}=-\frac{4}{3}\\ \\ x_2=\frac{2+10}{2.3}=\frac{12}{6}=2[/tex]
[tex]3x^2=2(x+4)\\
3x^2=2x+8\\
3x^2-2x-8=0\\
3x^2-6x+4x-8=0\\
3x(x-2)+4(x-2)=0\\
(3x+4)(x-2)=0\\
x=-\frac{4}{3} \vee x=2[/tex]
