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16t^2+27t+3=0
a = 16; b = 27; c = +3;
Δ = b2-4ac
Δ = 272-4·16·3
Δ = 537
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{537}}{2*16}=\frac{-27-\sqrt{537}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{537}}{2*16}=\frac{-27+\sqrt{537}}{32} $
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