165=w^2+4w

Solution for 165=w^2+4w equation:

165=w^2+4w

We move all terms to the left:

165-(w^2+4w)=0

We get rid of parentheses

-w^2-4w+165=0

We add all the numbers together, and all the variables

-1w^2-4w+165=0

a = -1; b = -4; c = +165;
Δ = b2-4ac
Δ = -42-4·(-1)·165
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-26}{2*-1}=\frac{-22}{-2}
=+11
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+26}{2*-1}=\frac{30}{-2}
=-15
$