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5t^2-13t+6=0
a = 5; b = -13; c = +6;
Δ = b2-4ac
Δ = -132-4·5·6
Δ = 49
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-7}{2*5}=\frac{6}{10} =3/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+7}{2*5}=\frac{20}{10} =2 $
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