0=2w^2+7w+3

Solution for 0=2w^2+7w+3 equation:

0=2w^2+7w+3

We move all terms to the left:

0-(2w^2+7w+3)=0

We add all the numbers together, and all the variables

-(2w^2+7w+3)=0

We get rid of parentheses

-2w^2-7w-3=0

a = -2; b = -7; c = -3;
Δ = b2-4ac
Δ = -72-4·(-2)·(-3)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*-2}=\frac{2}{-4}
=-1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*-2}=\frac{12}{-4}
=-3
$