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

Solution for (3+w)(4w-2)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4w^2-2w+12w-6)=0

We get rid of parentheses

4w^2-2w+12w-6=0

We add all the numbers together, and all the variables

4w^2+10w-6=0

a = 4; b = 10; c = -6;
Δ = b2-4ac
Δ = 102-4·4·(-6)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*4}=\frac{-24}{8}
=-3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*4}=\frac{4}{8}
=1/2
$