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

Solution for (5+w)(4w-2)=0 equation:

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

We add all the numbers together, and all the variables

(w+5)(4w-2)=0

We multiply parentheses ..

(+4w^2-2w+20w-10)=0

We get rid of parentheses

4w^2-2w+20w-10=0

We add all the numbers together, and all the variables

4w^2+18w-10=0

a = 4; b = 18; c = -10;
Δ = b2-4ac
Δ = 182-4·4·(-10)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*4}=\frac{-40}{8}
=-5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*4}=\frac{4}{8}
=1/2
$