3a+a=4a^2

Solution for 3a+a=4a^2 equation:

3a+a=4a^2

We move all terms to the left:

3a+a-(4a^2)=0

determiningTheFunctionDomain
-4a^2+3a+a=0

We add all the numbers together, and all the variables

-4a^2+4a=0

a = -4; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-4)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-4}=\frac{-8}{-8}
=1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-4}=\frac{0}{-8}
=0
$