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8t^2-15t-5=0
a = 8; b = -15; c = -5;
Δ = b2-4ac
Δ = -152-4·8·(-5)
Δ = 385
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{-(-15)-\sqrt{385}}{2*8}=\frac{15-\sqrt{385}}{16} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{385}}{2*8}=\frac{15+\sqrt{385}}{16} $
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