(2w+1)(4-w)=0

Solution for (2w+1)(4-w)=0 equation:

(2w+1)(4-w)=0

We add all the numbers together, and all the variables

(2w+1)(-1w+4)=0

We multiply parentheses ..

(-2w^2+8w-1w+4)=0

We get rid of parentheses

-2w^2+8w-1w+4=0

We add all the numbers together, and all the variables

-2w^2+7w+4=0

a = -2; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·(-2)·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{-(7)-9}{2*-2}=\frac{-16}{-4}
=+4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*-2}=\frac{2}{-4}
=-1/2
$