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4.9t^2+2t-6=0
a = 4.9; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·4.9·(-6)
Δ = 121.6
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{-(2)-\sqrt{121.6}}{2*4.9}=\frac{-2-\sqrt{121.6}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{121.6}}{2*4.9}=\frac{-2+\sqrt{121.6}}{9.8} $
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