t^2+2t+0.5=0

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Solution for t^2+2t+0.5=0 equation: 



t^2+2t+0.5=0

a = 1; b = 2; c = +0.5;
Δ = b2-4ac
Δ = 22-4·1·0.5
Δ = 2
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{-(2)-\sqrt{2}}{2*1}=\frac{-2-\sqrt{2}}{2}
$


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

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