4(a-3)=8a(4a+12)

Solution for 4(a-3)=8a(4a+12) equation:

4(a-3)=8a(4a+12)

We move all terms to the left:

4(a-3)-(8a(4a+12))=0

We multiply parentheses

4a-(8a(4a+12))-12=0


We calculate terms in parentheses: -(8a(4a+12)), so:

8a(4a+12)

We multiply parentheses

32a^2+96a

Back to the equation:

-(32a^2+96a)


We get rid of parentheses

-32a^2+4a-96a-12=0

We add all the numbers together, and all the variables

-32a^2-92a-12=0

a = -32; b = -92; c = -12;
Δ = b2-4ac
Δ = -922-4·(-32)·(-12)
Δ = 6928
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:

$\sqrt{\Delta}=\sqrt{6928}=\sqrt{16*433}=\sqrt{16}*\sqrt{433}=4\sqrt{433}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-92)-4\sqrt{433}}{2*-32}=\frac{92-4\sqrt{433}}{-64}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-92)+4\sqrt{433}}{2*-32}=\frac{92+4\sqrt{433}}{-64}
$