How do you solve #h²=32-4h#?

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#

#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)#