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=-16H^2+96H+5
We move all terms to the left:
-(-16H^2+96H+5)=0
We get rid of parentheses
16H^2-96H-5=0
a = 16; b = -96; c = -5;
Δ = b2-4ac
Δ = -962-4·16·(-5)
Δ = 9536
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9536}=\sqrt{64*149}=\sqrt{64}*\sqrt{149}=8\sqrt{149}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-8\sqrt{149}}{2*16}=\frac{96-8\sqrt{149}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+8\sqrt{149}}{2*16}=\frac{96+8\sqrt{149}}{32} $
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