(2d-5)(2d-4)=4d+d+20

Solution for (2d-5)(2d-4)=4d+d+20 equation:

(2d-5)(2d-4)=4d+d+20

We move all terms to the left:

(2d-5)(2d-4)-(4d+d+20)=0

We add all the numbers together, and all the variables

(2d-5)(2d-4)-(5d+20)=0

We get rid of parentheses

(2d-5)(2d-4)-5d-20=0

We multiply parentheses ..

(+4d^2-8d-10d+20)-5d-20=0

We get rid of parentheses

4d^2-8d-10d-5d+20-20=0

We add all the numbers together, and all the variables

4d^2-23d=0

a = 4; b = -23; c = 0;
Δ = b2-4ac
Δ = -232-4·4·0
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-23}{2*4}=\frac{0}{8}
=0
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+23}{2*4}=\frac{46}{8}
=5+3/4
$