Given:
The solution of quadratic equation is,
[tex]w=\frac{-1\pm\sqrt[]{10}}{3}[/tex]
So, the roots are written as,
[tex]\begin{gathered} w=\frac{-1+\sqrt[]{10}}{3},\frac{-1-\sqrt[]{10}}{3} \\ (x-(\frac{-1+\sqrt[]{10}}{3}))(x-(\frac{-1-\sqrt[]{10}}{3}))=0 \\ (x+\frac{1}{3}-\frac{\sqrt[]{10}}{3})(x+\frac{1}{3}+\frac{\sqrt[]{10}}{3})=0 \\ (x+\frac{1-\sqrt[]{10}}{3})(x+\frac{1+\sqrt[]{10}}{3})=0 \\ x^2+\frac{1+\sqrt[]{10}}{3}x+\frac{1-\sqrt[]{10}}{3}x+(\frac{1-\sqrt[]{10}}{3})(\frac{1+\sqrt[]{10}}{3})=0_{} \\ x^2+\frac{1}{3}x+\frac{\sqrt[]{10}}{3}x+\frac{1}{3}x-\frac{\sqrt[]{10}}{3}x+\frac{1-10}{9}=0 \\ x^2+\frac{2}{3}x-1=0 \end{gathered}[/tex]
Answer: The standard form quadratic equation is,
[tex]x^2+\frac{2}{3}x-1=0[/tex]
