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(H)=-25H^2-5H+1260
We move all terms to the left:
(H)-(-25H^2-5H+1260)=0
We get rid of parentheses
25H^2+5H+H-1260=0
We add all the numbers together, and all the variables
25H^2+6H-1260=0
a = 25; b = 6; c = -1260;
Δ = b2-4ac
Δ = 62-4·25·(-1260)
Δ = 126036
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{126036}=\sqrt{324*389}=\sqrt{324}*\sqrt{389}=18\sqrt{389}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18\sqrt{389}}{2*25}=\frac{-6-18\sqrt{389}}{50} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18\sqrt{389}}{2*25}=\frac{-6+18\sqrt{389}}{50} $
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