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=102-16H^2
We move all terms to the left:
-(102-16H^2)=0
We get rid of parentheses
16H^2-102=0
a = 16; b = 0; c = -102;
Δ = b2-4ac
Δ = 02-4·16·(-102)
Δ = 6528
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6528}=\sqrt{64*102}=\sqrt{64}*\sqrt{102}=8\sqrt{102}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{102}}{2*16}=\frac{0-8\sqrt{102}}{32} =-\frac{8\sqrt{102}}{32} =-\frac{\sqrt{102}}{4} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{102}}{2*16}=\frac{0+8\sqrt{102}}{32} =\frac{8\sqrt{102}}{32} =\frac{\sqrt{102}}{4} $
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