(w^2+120)+w+6w=180

Solution for (w^2+120)+w+6w=180 equation:

(w^2+120)+w+6w=180

We move all terms to the left:

(w^2+120)+w+6w-(180)=0

We add all the numbers together, and all the variables

7w+(w^2+120)-180=0

We get rid of parentheses

w^2+7w+120-180=0

We add all the numbers together, and all the variables

w^2+7w-60=0

a = 1; b = 7; c = -60;
Δ = b2-4ac
Δ = 72-4·1·(-60)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*1}=\frac{-24}{2}
=-12
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*1}=\frac{10}{2}
=5
$