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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4w^2+3w+16w+12)=0

We get rid of parentheses

4w^2+3w+16w+12=0

We add all the numbers together, and all the variables

4w^2+19w+12=0

a = 4; b = 19; c = +12;
Δ = b2-4ac
Δ = 192-4·4·12
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*4}=\frac{-32}{8}
=-4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*4}=\frac{-6}{8}
=-3/4
$