Question

The product of three and a squared number is twice the sum of the number and four

Answer 1

From the text of the task we can write the equation:


$3 x^{2} =2(x+4)\\ \\ 3x^2=2x+8\\ \\ 3x^2-2x-8=0\\ \\ \Delta=(-2)^2-4.3.(-8)=4+96=100\\ \\ x=\frac{2 \pm\sqrt{100}}{2.3}=\frac{2 \pm10}{6}\\ \\ x_1=\frac{2-10}{6}=-\frac{8}{6}=-\frac{4}{3}\\ \\ x_2=\frac{2+10}{2.3}=\frac{12}{6}=2$

Answer 2

$3x^2=2(x+4)\\ 3x^2=2x+8\\ 3x^2-2x-8=0\\ 3x^2-6x+4x-8=0\\ 3x(x-2)+4(x-2)=0\\ (3x+4)(x-2)=0\\ x=-\frac{4}{3} \vee x=2$