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6t^2-13t-15=0
a = 6; b = -13; c = -15;
Δ = b2-4ac
Δ = -132-4·6·(-15)
Δ = 529
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{529}=23$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-23}{2*6}=\frac{-10}{12} =-5/6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+23}{2*6}=\frac{36}{12} =3 $
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