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=-16H^2+210H-3
We move all terms to the left:
-(-16H^2+210H-3)=0
We get rid of parentheses
16H^2-210H+3=0
a = 16; b = -210; c = +3;
Δ = b2-4ac
Δ = -2102-4·16·3
Δ = 43908
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{43908}=\sqrt{4*10977}=\sqrt{4}*\sqrt{10977}=2\sqrt{10977}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-210)-2\sqrt{10977}}{2*16}=\frac{210-2\sqrt{10977}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-210)+2\sqrt{10977}}{2*16}=\frac{210+2\sqrt{10977}}{32} $
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