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4.9t^2-10t+2.5=0
a = 4.9; b = -10; c = +2.5;
Δ = b2-4ac
Δ = -102-4·4.9·2.5
Δ = 51
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{-(-10)-\sqrt{51}}{2*4.9}=\frac{10-\sqrt{51}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{51}}{2*4.9}=\frac{10+\sqrt{51}}{9.8} $
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