2t^2-5t-2=0

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



2t^2-5t-2=0

a = 2; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·2·(-2)
Δ = 41
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{41}}{2*2}=\frac{5-\sqrt{41}}{4}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{41}}{2*2}=\frac{5+\sqrt{41}}{4}
$

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