# How to solve for h?

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

##### 2 Answers

#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...

Feb 17, 2018