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=40+54H+-16H^2
We move all terms to the left:
-(40+54H+-16H^2)=0
We use the square of the difference formula
-(40+54H-16H^2)=0
We get rid of parentheses
16H^2-54H-40=0
a = 16; b = -54; c = -40;
Δ = b2-4ac
Δ = -542-4·16·(-40)
Δ = 5476
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{5476}=74$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-54)-74}{2*16}=\frac{-20}{32} =-5/8 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-54)+74}{2*16}=\frac{128}{32} =4 $
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