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16t^2+31t-5=0
a = 16; b = 31; c = -5;
Δ = b2-4ac
Δ = 312-4·16·(-5)
Δ = 1281
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{-(31)-\sqrt{1281}}{2*16}=\frac{-31-\sqrt{1281}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{1281}}{2*16}=\frac{-31+\sqrt{1281}}{32} $
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