3(4d+1)19d=6(2-d)

Solution for 3(4d+1)19d=6(2-d) equation:

3(4d+1)19d=6(2-d)

We move all terms to the left:

3(4d+1)19d-(6(2-d))=0

We add all the numbers together, and all the variables

3(4d+1)19d-(6(-1d+2))=0

We multiply parentheses

228d^2+57d-(6(-1d+2))=0


We calculate terms in parentheses: -(6(-1d+2)), so:

6(-1d+2)

We multiply parentheses

-6d+12

Back to the equation:

-(-6d+12)


We get rid of parentheses

228d^2+57d+6d-12=0

We add all the numbers together, and all the variables

228d^2+63d-12=0

a = 228; b = 63; c = -12;
Δ = b2-4ac
Δ = 632-4·228·(-12)
Δ = 14913
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:

$\sqrt{\Delta}=\sqrt{14913}=\sqrt{9*1657}=\sqrt{9}*\sqrt{1657}=3\sqrt{1657}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-3\sqrt{1657}}{2*228}=\frac{-63-3\sqrt{1657}}{456}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+3\sqrt{1657}}{2*228}=\frac{-63+3\sqrt{1657}}{456}
$