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

Solution for (2w+1)(w+4)=5 equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2w^2+9w-1=0

a = 2; b = 9; c = -1;
Δ = b2-4ac
Δ = 92-4·2·(-1)
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{89}}{2*2}=\frac{-9-\sqrt{89}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{89}}{2*2}=\frac{-9+\sqrt{89}}{4}
$