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-4.65t^2+20t+14=0
a = -4.65; b = 20; c = +14;
Δ = b2-4ac
Δ = 202-4·(-4.65)·14
Δ = 660.4
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{-(20)-\sqrt{660.4}}{2*-4.65}=\frac{-20-\sqrt{660.4}}{-9.3} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{660.4}}{2*-4.65}=\frac{-20+\sqrt{660.4}}{-9.3} $
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