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Sagot :
In order to solve the quadratic equation, first let's put it in the standard form:
[tex]\begin{gathered} y=ax^2+by+c \\ \\ 3x^2-6x=1 \\ 3x^2-6x-1=0 \\ \\ a=3,b=-6,c=-1 \end{gathered}[/tex]Then, let's use the quadratic formula to find the zeros of the equation:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ x_1=\frac{6+\sqrt[]{36+12}}{6}=\frac{6+\sqrt[]{48}}{6}=2.1547 \\ x_2=\frac{6-\sqrt[]{48}}{6}=-0.1547 \end{gathered}[/tex]So the solution of this quadratic equation is x = 2.1547 or x = -0.1547.
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