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(H)=-5H^2+40H+25
We move all terms to the left:
(H)-(-5H^2+40H+25)=0
We get rid of parentheses
5H^2-40H+H-25=0
We add all the numbers together, and all the variables
5H^2-39H-25=0
a = 5; b = -39; c = -25;
Δ = b2-4ac
Δ = -392-4·5·(-25)
Δ = 2021
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{2021}}{2*5}=\frac{39-\sqrt{2021}}{10} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{2021}}{2*5}=\frac{39+\sqrt{2021}}{10} $
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