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(H)=3H^2=1
We move all terms to the left:
(H)-(3H^2)=0
determiningTheFunctionDomain -3H^2+H=0
a = -3; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-3)·0
Δ = 1
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-3}=\frac{-2}{-6} =1/3 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-3}=\frac{0}{-6} =0 $
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