5t^2-4t-1=0

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Solution for 5t^2-4t-1=0 equation: 



5t^2-4t-1=0

a = 5; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·5·(-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{-(-4)-6}{2*5}=\frac{-2}{10}
=-1/5
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*5}=\frac{10}{10}
=1
$

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