3(a-4)=6a(3a+12)

Solution for 3(a-4)=6a(3a+12) equation:

3(a-4)=6a(3a+12)

We move all terms to the left:

3(a-4)-(6a(3a+12))=0

We multiply parentheses

3a-(6a(3a+12))-12=0


We calculate terms in parentheses: -(6a(3a+12)), so:

6a(3a+12)

We multiply parentheses

18a^2+72a

Back to the equation:

-(18a^2+72a)


We get rid of parentheses

-18a^2+3a-72a-12=0

We add all the numbers together, and all the variables

-18a^2-69a-12=0

a = -18; b = -69; c = -12;
Δ = b2-4ac
Δ = -692-4·(-18)·(-12)
Δ = 3897
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3897}=\sqrt{9*433}=\sqrt{9}*\sqrt{433}=3\sqrt{433}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-69)-3\sqrt{433}}{2*-18}=\frac{69-3\sqrt{433}}{-36}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-69)+3\sqrt{433}}{2*-18}=\frac{69+3\sqrt{433}}{-36}
$