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3t^2+13t-9=0
a = 3; b = 13; c = -9;
Δ = b2-4ac
Δ = 132-4·3·(-9)
Δ = 277
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{-(13)-\sqrt{277}}{2*3}=\frac{-13-\sqrt{277}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{277}}{2*3}=\frac{-13+\sqrt{277}}{6} $
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