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=40-5H-5H^2
We move all terms to the left:
-(40-5H-5H^2)=0
We get rid of parentheses
5H^2+5H-40=0
a = 5; b = 5; c = -40;
Δ = b2-4ac
Δ = 52-4·5·(-40)
Δ = 825
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{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{33}}{2*5}=\frac{-5-5\sqrt{33}}{10}H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{33}}{2*5}=\frac{-5+5\sqrt{33}}{10}
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