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(H)=100H^2
We move all terms to the left:
(H)-(100H^2)=0
determiningTheFunctionDomain -100H^2+H=0
a = -100; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-100)·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$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-100}=\frac{-2}{-200} =1/100 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-100}=\frac{0}{-200} =0 $
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