0=w^2+30w+161

Solution for 0=w^2+30w+161 equation:

0=w^2+30w+161

We move all terms to the left:

0-(w^2+30w+161)=0

We add all the numbers together, and all the variables

-(w^2+30w+161)=0

We get rid of parentheses

-w^2-30w-161=0

We add all the numbers together, and all the variables

-1w^2-30w-161=0

a = -1; b = -30; c = -161;
Δ = b2-4ac
Δ = -302-4·(-1)·(-161)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-16}{2*-1}=\frac{14}{-2}
=-7
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+16}{2*-1}=\frac{46}{-2}
=-23
$