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(3t)^2-5(3t)-14=0
a = 3; b = -53; c = -14;
Δ = b2-4ac
Δ = -532-4·3·(-14)
Δ = 2977
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{-(-53)-\sqrt{2977}}{2*3}=\frac{53-\sqrt{2977}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+\sqrt{2977}}{2*3}=\frac{53+\sqrt{2977}}{6} $
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