9(d-2)=3(d-12)d=

Solution for 9(d-2)=3(d-12)d= equation:

9(d-2)=3(d-12)d=

We move all terms to the left:

9(d-2)-(3(d-12)d)=0

We multiply parentheses

9d-(3(d-12)d)-18=0


We calculate terms in parentheses: -(3(d-12)d), so:

3(d-12)d

We multiply parentheses

3d^2-36d

Back to the equation:

-(3d^2-36d)


We get rid of parentheses

-3d^2+9d+36d-18=0

We add all the numbers together, and all the variables

-3d^2+45d-18=0

a = -3; b = 45; c = -18;
Δ = b2-4ac
Δ = 452-4·(-3)·(-18)
Δ = 1809
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:

$\sqrt{\Delta}=\sqrt{1809}=\sqrt{9*201}=\sqrt{9}*\sqrt{201}=3\sqrt{201}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-3\sqrt{201}}{2*-3}=\frac{-45-3\sqrt{201}}{-6}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+3\sqrt{201}}{2*-3}=\frac{-45+3\sqrt{201}}{-6}
$