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=-16H^2+16H+96
We move all terms to the left:
-(-16H^2+16H+96)=0
We get rid of parentheses
16H^2-16H-96=0
a = 16; b = -16; c = -96;
Δ = b2-4ac
Δ = -162-4·16·(-96)
Δ = 6400
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{6400}=80$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-80}{2*16}=\frac{-64}{32} =-2 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+80}{2*16}=\frac{96}{32} =3 $
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