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(H)=-2/3(H)-3
We move all terms to the left:
(H)-(-2/3(H)-3)=0
Domain of the equation: 3H-3)!=0We get rid of parentheses
H∈R
H+2/3H+3=0
We multiply all the terms by the denominator
H*3H+3*3H+2=0
Wy multiply elements
3H^2+9H+2=0
a = 3; b = 9; c = +2;
Δ = b2-4ac
Δ = 92-4·3·2
Δ = 57
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{-(9)-\sqrt{57}}{2*3}=\frac{-9-\sqrt{57}}{6} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{57}}{2*3}=\frac{-9+\sqrt{57}}{6} $
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