Given:
The quadratic equation is,
[tex]3x^2+5x-10=0[/tex]First find the discriminant to determine the nature of root,
[tex]\begin{gathered} ax^2+bx+c=0 \\ D=b^2-4ac \\ D=5^2-4(3)(-10) \\ =145>0 \\ So,\text{ the root are real and distinct. } \end{gathered}[/tex]Use the formula method,
[tex]\begin{gathered} 3x^2+5x-10=0 \\ a=3,b=5,c=-10 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-5\pm\sqrt{5^2-4\cdot\:3\left(-10\right)}}{2\cdot\:3} \\ x=\frac{-5\pm\sqrt{145}}{2\cdot\:3} \\ x=\frac{-5+\sqrt{145}}{2\cdot\:3},\: x_{}=\frac{-5-\sqrt{145}}{2\cdot\:3} \\ x=\frac{-5+\sqrt{145}}{6},\: x=\frac{-5-\sqrt{145}}{6} \end{gathered}[/tex]Answer: the root of the given quadratic equation is,
[tex]x=\frac{-5+\sqrt[]{145}}{6},\: x=\frac{-5-\sqrt[]{145}}{6}[/tex]