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=40H-2H^2
We move all terms to the left:
-(40H-2H^2)=0
We get rid of parentheses
2H^2-40H=0
a = 2; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·2·0
Δ = 1600
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{1600}=40$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*2}=\frac{0}{4} =0 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*2}=\frac{80}{4} =20 $
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