w2+28w+41=0

Solution for w2+28w+41=0 equation:

w2+28w+41=0

We add all the numbers together, and all the variables

w^2+28w+41=0

a = 1; b = 28; c = +41;
Δ = b2-4ac
Δ = 282-4·1·41
Δ = 620
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:

$\sqrt{\Delta}=\sqrt{620}=\sqrt{4*155}=\sqrt{4}*\sqrt{155}=2\sqrt{155}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{155}}{2*1}=\frac{-28-2\sqrt{155}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{155}}{2*1}=\frac{-28+2\sqrt{155}}{2}
$