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=16S-4.9S^2
We move all terms to the left:
-(16S-4.9S^2)=0
We get rid of parentheses
4.9S^2-16S=0
a = 4.9; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·4.9·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*4.9}=\frac{0}{9.8} =0 $$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*4.9}=\frac{32}{9.8} =3+2.6/9.8 $
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