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0.05t^2-2t-6.4=0
a = 0.05; b = -2; c = -6.4;
Δ = b2-4ac
Δ = -22-4·0.05·(-6.4)
Δ = 5.28
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{5.28}}{2*0.05}=\frac{2-\sqrt{5.28}}{0.1} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{5.28}}{2*0.05}=\frac{2+\sqrt{5.28}}{0.1} $
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