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8t^2+11t-164=0
a = 8; b = 11; c = -164;
Δ = b2-4ac
Δ = 112-4·8·(-164)
Δ = 5369
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{-(11)-\sqrt{5369}}{2*8}=\frac{-11-\sqrt{5369}}{16} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{5369}}{2*8}=\frac{-11+\sqrt{5369}}{16} $
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