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16t^2+16t-1=0
a = 16; b = 16; c = -1;
Δ = b2-4ac
Δ = 162-4·16·(-1)
Δ = 320
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{5}}{2*16}=\frac{-16-8\sqrt{5}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{5}}{2*16}=\frac{-16+8\sqrt{5}}{32} $
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