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4.9t^2+14.5t=0
a = 4.9; b = 14.5; c = 0;
Δ = b2-4ac
Δ = 14.52-4·4.9·0
Δ = 210.25
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14.5)-\sqrt{210.25}}{2*4.9}=\frac{-14.5-\sqrt{210.25}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14.5)+\sqrt{210.25}}{2*4.9}=\frac{-14.5+\sqrt{210.25}}{9.8} $
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