Y+t^2+2t=0

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



+Y^2+2Y=0

a = 1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·1·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*1}=\frac{-4}{2}
=-2
$


$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*1}=\frac{0}{2}
=0
$

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