Given:
The equation is:
[tex]\begin{gathered} 3x^2+6x-1=8 \\ 3x^2+6x-1-8=0 \\ 3x^2+6x-9=0 \end{gathered}[/tex]Take out common 3 from the equation.
[tex]\begin{gathered} 3(x^2+2x-3)=0 \\ x^2+2x-3=0 \end{gathered}[/tex]This is a quadratic equation.
The standard form of the quadratic equation is:
[tex]ax^2+bx+c=0[/tex]Then the solution of this equation is given as:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]The solution of the equation is:
[tex]\begin{gathered} x=\frac{-2\pm\sqrt{(2)^2-4(1)(-3)}}{2(1)} \\ x=\frac{-2\pm\sqrt{4+12}}{2} \\ x=\frac{-2\pm\sqrt{16}}{2} \\ x=\frac{-2\pm4}{2} \end{gathered}[/tex]Take + sign
[tex]\begin{gathered} x=\frac{-2+4}{2} \\ x=\frac{2}{2} \\ x=1 \end{gathered}[/tex]Take - sign
[tex]\begin{gathered} x=\frac{-2-4}{2} \\ x=\frac{-6}{2} \\ x=-3 \end{gathered}[/tex]Final Answer:
The solution of the given equation is x= -3, 1.
