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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+2w^2+8w+5w+20)=0

We get rid of parentheses

2w^2+8w+5w+20=0

We add all the numbers together, and all the variables

2w^2+13w+20=0

a = 2; b = 13; c = +20;
Δ = b2-4ac
Δ = 132-4·2·20
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*2}=\frac{-16}{4}
=-4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*2}=\frac{-10}{4}
=-2+1/2
$