Question

How do you solve `y^ { 2} - 12y + 19= 0`?

Answer 1

`x = 6 +- sqrt17`

Explanation:

`f(y) = y^2 - 12y + 19 = 0`
`D = d^2 = b^2 - 4ac = 144 - 76 = 68` --> `d = +- 2sqrt17`
There are 2 real roots:
`x = - b/(2a) +- d/(2a) = 12/2 +- 2sqrt17/2 = 6 +- sqrt17`