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8t^2-29t-2=0
a = 8; b = -29; c = -2;
Δ = b2-4ac
Δ = -292-4·8·(-2)
Δ = 905
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{-(-29)-\sqrt{905}}{2*8}=\frac{29-\sqrt{905}}{16} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{905}}{2*8}=\frac{29+\sqrt{905}}{16} $
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