# -4.9t^2+14t+98=0

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## Solution for -4.9t^2+14t+98=0 equation:

-4.9t^2+14t+98=0
a = -4.9; b = 14; c = +98;Δ = b2-4acΔ = 142-4·(-4.9)·98Δ = 2116.8The delta value is higher than zero, so the equation has two solutionsWe 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{-(14)-\sqrt{2116.8}}{2*-4.9}=\frac{-14-\sqrt{2116.8}}{-9.8}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+\sqrt{2116.8}}{2*-4.9}=\frac{-14+\sqrt{2116.8}}{-9.8}$

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