(2w+5)(w-8)=0

Solution for (2w+5)(w-8)=0 equation:

(2w+5)(w-8)=0

We multiply parentheses ..

(+2w^2-16w+5w-40)=0

We get rid of parentheses

2w^2-16w+5w-40=0

We add all the numbers together, and all the variables

2w^2-11w-40=0

a = 2; b = -11; c = -40;
Δ = b2-4ac
Δ = -112-4·2·(-40)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-21}{2*2}=\frac{-10}{4}
=-2+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+21}{2*2}=\frac{32}{4}
=8
$