(5w-4)(1+w)=0

Solution for (5w-4)(1+w)=0 equation:

(5w-4)(1+w)=0

We add all the numbers together, and all the variables

(5w-4)(w+1)=0

We multiply parentheses ..

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

We get rid of parentheses

5w^2+5w-4w-4=0

We add all the numbers together, and all the variables

5w^2+w-4=0

a = 5; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·5·(-4)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*5}=\frac{-10}{10}
=-1
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*5}=\frac{8}{10}
=4/5
$