s^2+3s=-1

Solution for s^2+3s=-1 equation:

s^2+3s=-1

We move all terms to the left:

s^2+3s-(-1)=0

We add all the numbers together, and all the variables

s^2+3s+1=0

a = 1; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·1·1
Δ = 5
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{5}}{2*1}=\frac{-3-\sqrt{5}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5}}{2*1}=\frac{-3+\sqrt{5}}{2}
$