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8t^2-2t-1=0
a = 8; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·8·(-1)
Δ = 36
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{36}=6$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*8}=\frac{-4}{16} =-1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*8}=\frac{8}{16} =1/2 $
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