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