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4.9t^2-39.2t-98=0
a = 4.9; b = -39.2; c = -98;
Δ = b2-4ac
Δ = -39.22-4·4.9·(-98)
Δ = 3457.44
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{-(-39.2)-\sqrt{3457.44}}{2*4.9}=\frac{39.2-\sqrt{3457.44}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39.2)+\sqrt{3457.44}}{2*4.9}=\frac{39.2+\sqrt{3457.44}}{9.8} $
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