(5w+8)(4w+2)=31w+20

Solution for (5w+8)(4w+2)=31w+20 equation:

(5w+8)(4w+2)=31w+20

We move all terms to the left:

(5w+8)(4w+2)-(31w+20)=0

We get rid of parentheses

(5w+8)(4w+2)-31w-20=0

We multiply parentheses ..

(+20w^2+10w+32w+16)-31w-20=0

We get rid of parentheses

20w^2+10w+32w-31w+16-20=0

We add all the numbers together, and all the variables

20w^2+11w-4=0

a = 20; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·20·(-4)
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*20}=\frac{-32}{40}
=-4/5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*20}=\frac{10}{40}
=1/4
$