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(H)=-H^2+2
We move all terms to the left:
(H)-(-H^2+2)=0
We get rid of parentheses
H^2+H-2=0
a = 1; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·1·(-2)
Δ = 9
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{9}=3$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*1}=\frac{-4}{2} =-2 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*1}=\frac{2}{2} =1 $
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