How do you solve #h^ { 2} - 2h - 0= 3#?

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Answer 1 · 2017-05-15T05:43:50

#h = 3, -1#

#h^2 - 2h - 0 = 3#

minus 3 to both sides,

#h^2 - 2h - 3 = 0#

use a quadratic formulae to solve where,
#a = 1, b = -2 and c =-3#

#h = (-b +-sqrt (b^2 - 4ac))/(2a)#

#h = ( -(-2) +- sqrt((-2^2) - 4(1)(-3)))/(2(1))#

#h = ( 2 +- sqrt(4+ 12))/2#

#h = ( 2 +- sqrt(16))/2 = (2 +- 4)/2#

#h = (2 +4)/2, (2 - 4)/2#

#h = 3, -1#