w^2-18w+69=0

Solution for w^2-18w+69=0 equation:

w^2-18w+69=0

a = 1; b = -18; c = +69;
Δ = b2-4ac
Δ = -182-4·1·69
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{3}}{2*1}=\frac{18-4\sqrt{3}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{3}}{2*1}=\frac{18+4\sqrt{3}}{2}
$