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6t^2-31t+35=0
a = 6; b = -31; c = +35;
Δ = b2-4ac
Δ = -312-4·6·35
Δ = 121
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{121}=11$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-11}{2*6}=\frac{20}{12} =1+2/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+11}{2*6}=\frac{42}{12} =3+1/2 $
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