Given:
[tex]3x^2+2x-8=0[/tex]Required:
To find the solution of the given equation.
Explanation:
Consider
[tex]3x^2+2x-8=0[/tex][tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex][tex]\begin{gathered} =\frac{-2\pm\sqrt{2^2-4(3)(-8)}}{2(3)} \\ \\ =\frac{-2\pm\sqrt{4+96}}{6} \\ \\ =\frac{-2\pm10}{6} \end{gathered}[/tex]Now
[tex]\begin{gathered} x=\frac{-2+10}{6} \\ \\ x=\frac{8}{6} \\ \\ x=\frac{4}{3} \end{gathered}[/tex]And
[tex]\begin{gathered} x=\frac{-2-10}{6} \\ \\ x=\frac{-12}{6} \\ \\ x=-2 \end{gathered}[/tex]Final Answer:
[tex]x=-2,\frac{4}{3}[/tex]