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=-4.9H^2+10H+1
We move all terms to the left:
-(-4.9H^2+10H+1)=0
We get rid of parentheses
4.9H^2-10H-1=0
a = 4.9; b = -10; c = -1;
Δ = b2-4ac
Δ = -102-4·4.9·(-1)
Δ = 119.6
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{-(-10)-\sqrt{119.6}}{2*4.9}=\frac{10-\sqrt{119.6}}{9.8} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{119.6}}{2*4.9}=\frac{10+\sqrt{119.6}}{9.8} $
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