How to solve for h?

#1/(h+1)+2=h#

2 Answers
Sunayan S · Jacobi J.
Feb 17, 2018

#1/(h+1)+2=h#

#=>(1+2(h+1))/(h+1)=h#

#=>3+2h=h(h+1)#

#=>3+2h=h^2+h#

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

Apply Shreedhar Acharya's Law (also known as the Quadratic Formula)

#h=(-(-1)+-sqrt((-1)^2-4 cdot 1 cdot (-3)))/2#
#=>h=(1+-sqrt13)/2#

hope it helps...
thank you...

Sahar Mulla ❤
Feb 17, 2018

#1/(1+h) + 2 = h#

#1/(1+h) = h-2#

#1= (h-2)(h+1)#
#1 = h^2 +h-2h-2#
#1= h^2-h-2#
#h^2-h-3=0#

#h=(-b+- sqrt(b^2-4ac))/(2a)#
#h=(1+- sqrt(1-4(-3)))/(2)#
#h=(1+- sqrt(13))/(2)#

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