s^2+3s-3=0

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Solution for s^2+3s-3=0 equation: 



s^2+3s-3=0

a = 1; b = 3; c = -3;
Δ = b2-4ac
Δ = 32-4·1·(-3)
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{21}}{2*1}=\frac{-3-\sqrt{21}}{2}
$


$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{21}}{2*1}=\frac{-3+\sqrt{21}}{2}
$

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