s^2-14s+23=0

Solution for s^2-14s+23=0 equation:

s^2-14s+23=0

a = 1; b = -14; c = +23;
Δ = b2-4ac
Δ = -142-4·1·23
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{26}}{2*1}=\frac{14-2\sqrt{26}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{26}}{2*1}=\frac{14+2\sqrt{26}}{2}
$