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

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

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

We add all the numbers together, and all the variables

(-1w+2)(4w+5)=0

We multiply parentheses ..

(-4w^2-5w+8w+10)=0

We get rid of parentheses

-4w^2-5w+8w+10=0

We add all the numbers together, and all the variables

-4w^2+3w+10=0

a = -4; b = 3; c = +10;
Δ = b2-4ac
Δ = 32-4·(-4)·10
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{-(3)-13}{2*-4}=\frac{-16}{-8}
=+2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*-4}=\frac{10}{-8}
=-1+1/4
$