0=3(2a)^2+7(2a)+8

Solution for 0=3(2a)^2+7(2a)+8 equation:

0=3(2a)^2+7(2a)+8

We move all terms to the left:

0-(3(2a)^2+7(2a)+8)=0

We add all the numbers together, and all the variables

-(32a^2+72a+8)=0

We get rid of parentheses

-32a^2-72a-8=0

a = -32; b = -72; c = -8;
Δ = b2-4ac
Δ = -722-4·(-32)·(-8)
Δ = 4160
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:

$\sqrt{\Delta}=\sqrt{4160}=\sqrt{64*65}=\sqrt{64}*\sqrt{65}=8\sqrt{65}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-8\sqrt{65}}{2*-32}=\frac{72-8\sqrt{65}}{-64}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+8\sqrt{65}}{2*-32}=\frac{72+8\sqrt{65}}{-64}
$