1/3a2+12=39

Solution for 1/3a2+12=39 equation:

1/3a^2+12=39

We move all terms to the left:

1/3a^2+12-(39)=0


Domain of the equation: 3a^2!=0

a^2!=0/3

a^2!=√0

a!=0

a∈R


We add all the numbers together, and all the variables

1/3a^2-27=0

We multiply all the terms by the denominator


-27*3a^2+1=0

Wy multiply elements

-81a^2+1=0

a = -81; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-81)·1
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*-81}=\frac{-18}{-162}
=1/9
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*-81}=\frac{18}{-162}
=-1/9
$