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4.9t^2+14t-39=0
a = 4.9; b = 14; c = -39;
Δ = b2-4ac
Δ = 142-4·4.9·(-39)
Δ = 960.4
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{-(14)-\sqrt{960.4}}{2*4.9}=\frac{-14-\sqrt{960.4}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+\sqrt{960.4}}{2*4.9}=\frac{-14+\sqrt{960.4}}{9.8} $
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