0=3w^2+13w+4

Solution for 0=3w^2+13w+4 equation:

0=3w^2+13w+4

We move all terms to the left:

0-(3w^2+13w+4)=0

We add all the numbers together, and all the variables

-(3w^2+13w+4)=0

We get rid of parentheses

-3w^2-13w-4=0

a = -3; b = -13; c = -4;
Δ = b2-4ac
Δ = -132-4·(-3)·(-4)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*-3}=\frac{2}{-6}
=-1/3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*-3}=\frac{24}{-6}
=-4
$