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125+5.2t-4.9t^2=0
a = -4.9; b = 5.2; c = +125;
Δ = b2-4ac
Δ = 5.22-4·(-4.9)·125
Δ = 2477.04
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{-(5.2)-\sqrt{2477.04}}{2*-4.9}=\frac{-5.2-\sqrt{2477.04}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.2)+\sqrt{2477.04}}{2*-4.9}=\frac{-5.2+\sqrt{2477.04}}{-9.8} $
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