Question

I need to solve this quadratic equation even if the solution with involves imaginary numbers

Answer 1

Given:

The quadratic equation is:

`25x^2+20x-59=0`

To solve the equation

`ax^2+bx+c=0`

we have its roots as:

`x=\text{ }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

According to our question, we have a= 25 b= 20 c= - 59

Thus,

`x=\frac{-20\pm\sqrt[]{20^2-4(25*-59)}}{2*25}`
`\begin{gathered} x=\frac{-20\pm\sqrt[]{400+5900}}{50} \\ =\frac{-20\pm\sqrt[]{6300}}{50} \end{gathered}`
`\begin{gathered} x=\frac{-20\pm30\sqrt[]{7}}{50} \\ =\frac{-2\pm3\sqrt[]{7}}{5} \\ =\frac{-2+3\sqrt[]{7}}{5},\frac{-2-3\sqrt[]{7}}{5} \end{gathered}`
`\begin{gathered} x=(-2+3(2.64575))/(5),(-2-3(2.64575))/(5) \\ x=(-2+7.93725)/(5),(-2-7.93725)/(5) \\ =(5.93725)/(5),(9.93725)/(5) \\ =1.18745,\text{ 1.98745} \end{gathered}`

Hence, x= 1.18745 , 1.98745