Question

How do you solve `2x^2 - 12x + 7 = 5`?

Answer 1

`x = 3+-2sqrt(2)`

Explanation:

Given:

`2x^2-12x+7=5`

Subtract `5` from both sides to get:

`2x^2-12x+2=0`

Divide both sides by `2` to get:

`0 = x^2-6x+1`

`color(white)(0) = x^2-6x+9-8`

`color(white)(0) = (x-3)^2-(2sqrt(2))^2`

`color(white)(0) = ((x-3)-2sqrt(2))((x-3)+2sqrt(2))`

`color(white)(0) = (x-3-2sqrt(2))(x-3+2sqrt(2))`

Hence:

`x = 3+-2sqrt(2)`

Answer 2

`x = 3 +- 2sqrt2`

Explanation:

2x^2 - 12x + 2 = 0
x^2 - 6x + 1 = 0
Use the improved quadratic formula (Socratic search):
`D = d^2 = b^2 - 4ac = 36 - 4 = 32 = 2(16)`
`d = +- 4sqrt2`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = - (-6/2) +- (4sqrt2)/2`
`x = (3 +- 2sqrt2)`