Question

How do you use the quadratic formula to solve `1/2x^2-3x+2=3x-1`?

Answer 1

`x_1~~11.477`
`x_2~~0.523`

Explanation:

Making right side of the equation equate as 0, we end up with:
`1/2x^2-6x+3=0`

We can make it easier to visualise by multiplying both sides by 2,
`x^2-12x+6=0`

How many and what kinds of roots will we end up with?
`b^2-4ac=(-12)^2-4(1)(6)=120 > 0` (i.e. 2 real roots)

Let's find out what they are:

`x_(1,2)=-b/(2a)+-sqrt(b^2-4ac)/(2a)`

`x_1=-(-12)/(2*1)+sqrt(120)/(2*1)~~11.477`
`x_2=-(-12)/(2*1)-sqrt(120)/(2*1)~~0.523`