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15t^2-2t-1=0
a = 15; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·15·(-1)
Δ = 64
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*15}=\frac{-6}{30} =-1/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*15}=\frac{10}{30} =1/3 $
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