s^2+10s+13=0

Solution for s^2+10s+13=0 equation:

s^2+10s+13=0

a = 1; b = 10; c = +13;
Δ = b2-4ac
Δ = 102-4·1·13
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{3}}{2*1}=\frac{-10-4\sqrt{3}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{3}}{2*1}=\frac{-10+4\sqrt{3}}{2}
$