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4.9t^2-8t+2=0
a = 4.9; b = -8; c = +2;
Δ = b2-4ac
Δ = -82-4·4.9·2
Δ = 24.8
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{-(-8)-\sqrt{24.8}}{2*4.9}=\frac{8-\sqrt{24.8}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+\sqrt{24.8}}{2*4.9}=\frac{8+\sqrt{24.8}}{9.8} $
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