a^2=(3a^2)+6a+-1224

Solution for a^2=(3a^2)+6a+-1224 equation:

a^2=(3a^2)+6a+-1224

We move all terms to the left:

a^2-((3a^2)+6a+-1224)=0

We use the square of the difference formula

a^2-(3a^2+6a-1224)=0

We get rid of parentheses

a^2-3a^2-6a+1224=0

We add all the numbers together, and all the variables

-2a^2-6a+1224=0

a = -2; b = -6; c = +1224;
Δ = b2-4ac
Δ = -62-4·(-2)·1224
Δ = 9828
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:

$\sqrt{\Delta}=\sqrt{9828}=\sqrt{36*273}=\sqrt{36}*\sqrt{273}=6\sqrt{273}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{273}}{2*-2}=\frac{6-6\sqrt{273}}{-4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{273}}{2*-2}=\frac{6+6\sqrt{273}}{-4}
$