s^2+4s=1

Solution for s^2+4s=1 equation:

s^2+4s=1

We move all terms to the left:

s^2+4s-(1)=0

a = 1; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·1·(-1)
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{5}}{2*1}=\frac{-4-2\sqrt{5}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{5}}{2*1}=\frac{-4+2\sqrt{5}}{2}
$