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(H)=1/3H+11
We move all terms to the left:
(H)-(1/3H+11)=0
Domain of the equation: 3H+11)!=0We get rid of parentheses
H∈R
H-1/3H-11=0
We multiply all the terms by the denominator
H*3H-11*3H-1=0
Wy multiply elements
3H^2-33H-1=0
a = 3; b = -33; c = -1;
Δ = b2-4ac
Δ = -332-4·3·(-1)
Δ = 1101
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-\sqrt{1101}}{2*3}=\frac{33-\sqrt{1101}}{6} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{1101}}{2*3}=\frac{33+\sqrt{1101}}{6} $
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