0=2w^2+5w-42

Solution for 0=2w^2+5w-42 equation:

0=2w^2+5w-42

We move all terms to the left:

0-(2w^2+5w-42)=0

We add all the numbers together, and all the variables

-(2w^2+5w-42)=0

We get rid of parentheses

-2w^2-5w+42=0

a = -2; b = -5; c = +42;
Δ = b2-4ac
Δ = -52-4·(-2)·42
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*-2}=\frac{-14}{-4}
=3+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*-2}=\frac{24}{-4}
=-6
$