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4.9t^2-9t+3=0
a = 4.9; b = -9; c = +3;
Δ = b2-4ac
Δ = -92-4·4.9·3
Δ = 22.2
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{-(-9)-\sqrt{22.2}}{2*4.9}=\frac{9-\sqrt{22.2}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{22.2}}{2*4.9}=\frac{9+\sqrt{22.2}}{9.8} $
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