t^2+t-0.25=0

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Solution for t^2+t-0.25=0 equation: 



t^2+t-0.25=0

a = 1; b = 1; c = -0.25;
Δ = b2-4ac
Δ = 12-4·1·(-0.25)
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{-(1)-\sqrt{2}}{2*1}=\frac{-1-\sqrt{2}}{2}
$


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

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