Given the equation:
[tex]6x^2+7x+2=0[/tex]
We will use the following rule to find the solution to the equation:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
From the given equation: a = 6, b = 7, c = 2
So,
[tex]\begin{gathered} x=\frac{-7\pm\sqrt[]{7^2-4\cdot6\cdot2}}{2\cdot6}=\frac{-7\pm\sqrt[]{1}}{12}=\frac{-7\pm1}{12} \\ x=\frac{-7-1}{12}=-\frac{8}{12}=-\frac{2}{3} \\ or,x=\frac{-7+1}{12}=-\frac{6}{12}=-\frac{1}{2} \end{gathered}[/tex]
So, the answer will be option B) x = -1/2, -2/3
