Question

How do you solve `6+ 3x = 8x^{2}`?

Answer 1

Solution is `3/16-sqrt201/16` and `3/16+sqrt201/16`

Explanation:

Transposing all the terms to RHS `6+3x=8x^2`

`hArr0=8x^2-3x-6` or `8x^2-3x-6=0`

This is general form of quadratic equation i.e. of type `ax^2+bx+c=0` and discriminant is

`b^2-4ac=(-3)^2-4xx8xx(-6)=9+192=201`, which is not a perfect square of a rational number and hence we cannot use splitting the middle term to solve this. So we use quadratic formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Hence `x=(-(-3)+-sqrt201)/(2xx8)`

= `(3+-sqrt201)/16`

Hence solution is `3/16-sqrt201/16` and `3/16+sqrt201/16`