11d(9d-2)=5d+28

Solution for 11d(9d-2)=5d+28 equation:

11d(9d-2)=5d+28

We move all terms to the left:

11d(9d-2)-(5d+28)=0

We multiply parentheses

99d^2-22d-(5d+28)=0

We get rid of parentheses

99d^2-22d-5d-28=0

We add all the numbers together, and all the variables

99d^2-27d-28=0

a = 99; b = -27; c = -28;
Δ = b2-4ac
Δ = -272-4·99·(-28)
Δ = 11817
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{11817}=\sqrt{9*1313}=\sqrt{9}*\sqrt{1313}=3\sqrt{1313}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3\sqrt{1313}}{2*99}=\frac{27-3\sqrt{1313}}{198}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3\sqrt{1313}}{2*99}=\frac{27+3\sqrt{1313}}{198}
$