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16t^2-27t-2=0
a = 16; b = -27; c = -2;
Δ = b2-4ac
Δ = -272-4·16·(-2)
Δ = 857
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{-(-27)-\sqrt{857}}{2*16}=\frac{27-\sqrt{857}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{857}}{2*16}=\frac{27+\sqrt{857}}{32} $
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