Solve for `x`

:

`(x - `

`H`)^2 + `K` = 0

`x = {}`

`\space \text{or} \space x = {}`

integers, like `6`

*simplified proper* fractions, like `3/5`

*simplified improper* fractions, like `7/4`

and/or *exact* decimals, like `0.75`

Add

to both sides so we can start isolating `abs( K )``x`

on the left:

`\qquad (x - `

`H`)^2 = `-K`

Take the square root of both sides to get rid of the exponent.

`\qquad \sqrt{(x - `

`H`)^2} = \pm \sqrt{`-K`}

Be sure to consider both positive and negative

, since squaring either one results
in `CONSTANT`

.
`-K`

`\qquad x - `

`H` = \pm `CONSTANT`

Add

to both sides to isolate `abs( H )``x`

on the left:
Subtract

from both sides to isolate `abs( H )``x`

on the left:

`\qquad x = `

`H` \pm `CONSTANT`

Add and subtract

to find the two possible solutions:`CONSTANT`

`\qquad x = `

`H + CONSTANT` \quad \text{or} \quad x = `H - CONSTANT`

Determine where `f(x)`

intersects the `x`

-axis.

`f(x) = (x - `

`H`)^2 + `K`

The function intersects the `x`

-axis where `f(x) = 0`

, so solve the equation:

`\qquad (x - `

`H`)^2 + `K` = 0