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16t^2-19t-3=0
a = 16; b = -19; c = -3;
Δ = b2-4ac
Δ = -192-4·16·(-3)
Δ = 553
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{-(-19)-\sqrt{553}}{2*16}=\frac{19-\sqrt{553}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{553}}{2*16}=\frac{19+\sqrt{553}}{32} $
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