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=0.8+29.4H-4.9H^2
We move all terms to the left:
-(0.8+29.4H-4.9H^2)=0
We get rid of parentheses
4.9H^2-29.4H-0.8=0
a = 4.9; b = -29.4; c = -0.8;
Δ = b2-4ac
Δ = -29.42-4·4.9·(-0.8)
Δ = 880.04
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{-(-29.4)-\sqrt{880.04}}{2*4.9}=\frac{29.4-\sqrt{880.04}}{9.8} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29.4)+\sqrt{880.04}}{2*4.9}=\frac{29.4+\sqrt{880.04}}{9.8} $
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