t^2-5t-1=0

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Solution for t^2-5t-1=0 equation: 



t^2-5t-1=0

a = 1; b = -5; c = -1;
Δ = b2-4ac
Δ = -52-4·1·(-1)
Δ = 29
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{-(-5)-\sqrt{29}}{2*1}=\frac{5-\sqrt{29}}{2}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{29}}{2*1}=\frac{5+\sqrt{29}}{2}
$

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