s^2+10s=-23

Solution for s^2+10s=-23 equation:

s^2+10s=-23

We move all terms to the left:

s^2+10s-(-23)=0

We add all the numbers together, and all the variables

s^2+10s+23=0

a = 1; b = 10; c = +23;
Δ = b2-4ac
Δ = 102-4·1·23
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{2}}{2*1}=\frac{-10-2\sqrt{2}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{2}}{2*1}=\frac{-10+2\sqrt{2}}{2}
$