3a^2+9a+1=5

Solution for 3a^2+9a+1=5 equation:

3a^2+9a+1=5

We move all terms to the left:

3a^2+9a+1-(5)=0

We add all the numbers together, and all the variables

3a^2+9a-4=0

a = 3; b = 9; c = -4;
Δ = b2-4ac
Δ = 92-4·3·(-4)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{129}}{2*3}=\frac{-9-\sqrt{129}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{129}}{2*3}=\frac{-9+\sqrt{129}}{6}
$