(2w+5)(2w+7)=28

Solution for (2w+5)(2w+7)=28 equation:

(2w+5)(2w+7)=28

We move all terms to the left:

(2w+5)(2w+7)-(28)=0

We multiply parentheses ..

(+4w^2+14w+10w+35)-28=0

We get rid of parentheses

4w^2+14w+10w+35-28=0

We add all the numbers together, and all the variables

4w^2+24w+7=0

a = 4; b = 24; c = +7;
Δ = b2-4ac
Δ = 242-4·4·7
Δ = 464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{29}}{2*4}=\frac{-24-4\sqrt{29}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{29}}{2*4}=\frac{-24+4\sqrt{29}}{8}
$