How do you solve #h^2=-3h+54#?
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Rearrange to get
#h^2 + 3h - 54 = 0#
Factorise
#( h+ 9)(h - 6) = 0#
#h + 9 = 0 or h- 6 = 0#
#h = - 9 or h =6#
#"arrange the equation in standard form"#
#rArrh^2+3h-54=0larrcolor(blue)"in standard form"#
#"the factors of - 54 which sum to + 3 are + 9 and - 6"#
#rArr(h+9)(h-6)=0#
#"equate each factor to zero and solve for h"#
#h+9=0rArrh=-9#
#h-6=0rArrh=6#
We have,
#color(white)(xxx)h^2 = -3h + 54#
#rArr h^2 + 3h - 54 = cancel(-3h) cancel(+ 54) cancel(+ 3h) cancel(- 54)# [Add #3h - 54# to both sides]
#rArr h^2 + 3h - 54 = 0#
#rArr h^2 + (9 - 6)h - 54 = 0# [Break #3# as #9 - 6#]
#rArrh^2 + 9h - 6h - 54 = 0#
#rArr h(h + 9) - 6(h + 9)# [Group Like Terms]
#rArr (h + 9)(h - 6) = 0# [Group Again]
So, If,
#h+ 9 = 0 rArr h = -9#
and if,
#h - 6 = 0 rArr h = 6#
Hence Explained.
