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4t+4.9t^2=0
a = 4.9; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·4.9·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*4.9}=\frac{-8}{9.8} =-8/9.8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*4.9}=\frac{0}{9.8} =0 $
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