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5t^2+4.9t-300=0
a = 5; b = 4.9; c = -300;
Δ = b2-4ac
Δ = 4.92-4·5·(-300)
Δ = 6024.01
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{-(4.9)-\sqrt{6024.01}}{2*5}=\frac{-4.9-\sqrt{6024.01}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.9)+\sqrt{6024.01}}{2*5}=\frac{-4.9+\sqrt{6024.01}}{10} $
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