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=-16H^2+24
We move all terms to the left:
-(-16H^2+24)=0
We get rid of parentheses
16H^2-24=0
a = 16; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·16·(-24)
Δ = 1536
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{6}}{2*16}=\frac{0-16\sqrt{6}}{32} =-\frac{16\sqrt{6}}{32} =-\frac{\sqrt{6}}{2} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{6}}{2*16}=\frac{0+16\sqrt{6}}{32} =\frac{16\sqrt{6}}{32} =\frac{\sqrt{6}}{2} $
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