w^2-12w+23=0

Solution for w^2-12w+23=0 equation:

w^2-12w+23=0

a = 1; b = -12; c = +23;
Δ = b2-4ac
Δ = -122-4·1·23
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{13}}{2*1}=\frac{12-2\sqrt{13}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{13}}{2*1}=\frac{12+2\sqrt{13}}{2}
$