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(H)=-16H^2+40H+200
We move all terms to the left:
(H)-(-16H^2+40H+200)=0
We get rid of parentheses
16H^2-40H+H-200=0
We add all the numbers together, and all the variables
16H^2-39H-200=0
a = 16; b = -39; c = -200;
Δ = b2-4ac
Δ = -392-4·16·(-200)
Δ = 14321
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{-(-39)-\sqrt{14321}}{2*16}=\frac{39-\sqrt{14321}}{32} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{14321}}{2*16}=\frac{39+\sqrt{14321}}{32} $
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