Question

How do you find the zeros of `y = 4/5x^2 - 7/2x + 2/3` using the quadratic formula?

Answer 1

`x_1=35/16+sqrt(9195)/48` or `x_2=35/16-sqrt(9105)/48`

Explanation:

Multiplying the whole equation by `30`
`24x^2-105x+20=0`
then we get
`x^2-105/24x+5/6=0`
`x_{1,2}=35/16pmsqrt((35/16)^2-5/6)`
`x_{1,2}=35/16pm\sqrt(3035/768)`
`x_{1,2}=35/16pm\sqrt(9105)/48`