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(3t)^2-3(5t)-14=0
a = 3; b = -35; c = -14;
Δ = b2-4ac
Δ = -352-4·3·(-14)
Δ = 1393
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{-(-35)-\sqrt{1393}}{2*3}=\frac{35-\sqrt{1393}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1393}}{2*3}=\frac{35+\sqrt{1393}}{6} $
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