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0.6t^2-4t+6=0
a = 0.6; b = -4; c = +6;
Δ = b2-4ac
Δ = -42-4·0.6·6
Δ = 1.6
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{-(-4)-\sqrt{1.6}}{2*0.6}=\frac{4-\sqrt{1.6}}{1.2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{1.6}}{2*0.6}=\frac{4+\sqrt{1.6}}{1.2} $
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