0=w^2+1w-56

Solution for 0=w^2+1w-56 equation:

0=w^2+1w-56

We move all terms to the left:

0-(w^2+1w-56)=0

We add all the numbers together, and all the variables

-(w^2+1w-56)=0

We get rid of parentheses

-w^2-1w+56=0

We add all the numbers together, and all the variables

-1w^2-1w+56=0

a = -1; b = -1; c = +56;
Δ = b2-4ac
Δ = -12-4·(-1)·56
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-15}{2*-1}=\frac{-14}{-2}
=+7
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+15}{2*-1}=\frac{16}{-2}
=-8
$