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6.1t^2+51t+100=0
a = 6.1; b = 51; c = +100;
Δ = b2-4ac
Δ = 512-4·6.1·100
Δ = 161
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{-(51)-\sqrt{161}}{2*6.1}=\frac{-51-\sqrt{161}}{12.2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+\sqrt{161}}{2*6.1}=\frac{-51+\sqrt{161}}{12.2} $
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