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4.9t^2-25t+5=0
a = 4.9; b = -25; c = +5;
Δ = b2-4ac
Δ = -252-4·4.9·5
Δ = 527
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{-(-25)-\sqrt{527}}{2*4.9}=\frac{25-\sqrt{527}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{527}}{2*4.9}=\frac{25+\sqrt{527}}{9.8} $
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