0=w^2+2w-35

Solution for 0=w^2+2w-35 equation:

0=w^2+2w-35

We move all terms to the left:

0-(w^2+2w-35)=0

We add all the numbers together, and all the variables

-(w^2+2w-35)=0

We get rid of parentheses

-w^2-2w+35=0

We add all the numbers together, and all the variables

-1w^2-2w+35=0

a = -1; b = -2; c = +35;
Δ = b2-4ac
Δ = -22-4·(-1)·35
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-12}{2*-1}=\frac{-10}{-2}
=+5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+12}{2*-1}=\frac{14}{-2}
=-7
$