Question

The solutions of the quadratic equation: 3x^2+5x-10=0Are x= _______Note: give your answer as a list of complex numbers, such as 3-4i, 5+i.

Answer 1

Given:

The quadratic equation is,

`3x^2+5x-10=0`

First find the discriminant to determine the nature of root,

`\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}`

Use the formula method,

`\begin{gathered} 3x^2+5x-10=0 \\ a=3,b=5,c=-10 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=(-5\pm√(5^2-4\cdot\:3\left(-10\right)))/(2\cdot\:3) \\ x=(-5\pm√(145))/(2\cdot\:3) \\ x=(-5+√(145))/(2\cdot\:3),\: x_{}=(-5-√(145))/(2\cdot\:3) \\ x=(-5+√(145))/(6),\: x=(-5-√(145))/(6) \end{gathered}`

Answer: the root of the given quadratic equation is,

`x=\frac{-5+\sqrt[]{145}}{6},\: x=\frac{-5-\sqrt[]{145}}{6}`