Question

How do you solve `h²=32-4h`?

Answer 1

Convert to standard form then either factor or use the quadratic formula to get
`color(white)("XXX")h=-8 or h=4`

Explanation:

Given
`color(white)("XXX")h^2=32-4h`

Re-write into standard form:
`color(white)("XXX")color(red)((1)h^2color(blue)(+4)hcolor(green)(-32) = 0`

Option 1
Recognize the factoring:
`color(white)("XXX")(h+8)(h-4)=0`
`rarrcolor(white)("XXX")h=-8 or h=4`

Option 2
Apply the quadratic formula for roots:
`color(white)("XXX")h=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2color(red)(a))`
in this specific case:
`color(white)("XXX")h=(-color(blue)(4)+-sqrt((color(blue)(4))^2-4(color(red)(1))(color(green)(-32))))/(2(color(red)(1)))`

`color(white)("XXX")=(-4+-sqrt(16+128))/2`

`color(white)("XXX")=(-4+-sqrt(144))/2`

`color(white)("XXX")=-2+-6`

`rarrcolor(white)("XXX")h=+4 or h=-8`

Answer 2

`h=4,-8`

Explanation:

`h^2=32-4h`

Gather all terms to one side of the equation and arrange the equation in standard form.

`h^2+4h-32=0`

Determine two numbers that when added equal `4` and when multiplied equal `-32`. The numbers `8` and `-4`.

Rewrite the equation in factored form.

`color(red)((h-4))color(blue)((h+8))=0`

Set each binomial equal to zero and solve for `h`.

`color(red)(h-4)=0`

`color(red)(h=4)`

`color(blue)(h+8)=0`

`color(blue)(h=-8)`

`h=color(red)4, color(blue)(-8)`