0=w^2-33w+161

Solution for 0=w^2-33w+161 equation:

0=w^2-33w+161

We move all terms to the left:

0-(w^2-33w+161)=0

We add all the numbers together, and all the variables

-(w^2-33w+161)=0

We get rid of parentheses

-w^2+33w-161=0

We add all the numbers together, and all the variables

-1w^2+33w-161=0

a = -1; b = 33; c = -161;
Δ = b2-4ac
Δ = 332-4·(-1)·(-161)
Δ = 445
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-\sqrt{445}}{2*-1}=\frac{-33-\sqrt{445}}{-2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{445}}{2*-1}=\frac{-33+\sqrt{445}}{-2}
$