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5-5t+0.83t^2=0
a = 0.83; b = -5; c = +5;
Δ = b2-4ac
Δ = -52-4·0.83·5
Δ = 8.4
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{-(-5)-\sqrt{8.4}}{2*0.83}=\frac{5-\sqrt{8.4}}{1.66} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{8.4}}{2*0.83}=\frac{5+\sqrt{8.4}}{1.66} $
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