Question

How do you solve `2x^2 + 32x + 12 = 0` using completing the square?

Answer 1

I found:
`x_1=-8+sqrt(58)`
`x_2=-8-sqrt(58)`

Explanation:

Write it as (dividing by `2`):
`x^2/2+32/2x+12/2=0`
`x^2+16x+6=0`
`x^2+16x=-6`
Add and subtract `64`:
`x^2+16x+64-64=-6`
`x^2+16x+64=-6+64`
`(x+8)^2=58`
`x+8=+-sqrt(58)`
`x=-8+-sqrt(58)`
so:
`x_1=-8+sqrt(58)`
`x_2=-8-sqrt(58)`