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=4+33H+-16H^2
We move all terms to the left:
-(4+33H+-16H^2)=0
We use the square of the difference formula
-(4+33H-16H^2)=0
We get rid of parentheses
16H^2-33H-4=0
a = 16; b = -33; c = -4;
Δ = b2-4ac
Δ = -332-4·16·(-4)
Δ = 1345
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}H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-\sqrt{1345}}{2*16}=\frac{33-\sqrt{1345}}{32}H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{1345}}{2*16}=\frac{33+\sqrt{1345}}{32}
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