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=-16H^2+227H+31
We move all terms to the left:
-(-16H^2+227H+31)=0
We get rid of parentheses
16H^2-227H-31=0
a = 16; b = -227; c = -31;
Δ = b2-4ac
Δ = -2272-4·16·(-31)
Δ = 53513
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-227)-\sqrt{53513}}{2*16}=\frac{227-\sqrt{53513}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-227)+\sqrt{53513}}{2*16}=\frac{227+\sqrt{53513}}{32} $
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