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.024t^2-0.578t+2.309=0
a = .024; b = -0.578; c = +2.309;
Δ = b2-4ac
Δ = -0.5782-4·.024·2.309
Δ = 0.11242
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{-(-0.578)-\sqrt{0.11242}}{2*.024}=\frac{0.578-\sqrt{0.11242}}{0.048} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.578)+\sqrt{0.11242}}{2*.024}=\frac{0.578+\sqrt{0.11242}}{0.048} $
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