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4.9t^2-29t+25=0
a = 4.9; b = -29; c = +25;
Δ = b2-4ac
Δ = -292-4·4.9·25
Δ = 351
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{-(-29)-\sqrt{351}}{2*4.9}=\frac{29-\sqrt{351}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{351}}{2*4.9}=\frac{29+\sqrt{351}}{9.8} $
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