(4+w)(3w+5)=0

Solution for (4+w)(3w+5)=0 equation:

(4+w)(3w+5)=0

We add all the numbers together, and all the variables

(w+4)(3w+5)=0

We multiply parentheses ..

(+3w^2+5w+12w+20)=0

We get rid of parentheses

3w^2+5w+12w+20=0

We add all the numbers together, and all the variables

3w^2+17w+20=0

a = 3; b = 17; c = +20;
Δ = b2-4ac
Δ = 172-4·3·20
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*3}=\frac{-24}{6}
=-4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*3}=\frac{-10}{6}
=-1+2/3
$