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=-16H^2+32H+24
We move all terms to the left:
-(-16H^2+32H+24)=0
We get rid of parentheses
16H^2-32H-24=0
a = 16; b = -32; c = -24;
Δ = b2-4ac
Δ = -322-4·16·(-24)
Δ = 2560
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{10}}{2*16}=\frac{32-16\sqrt{10}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{10}}{2*16}=\frac{32+16\sqrt{10}}{32} $
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