3a2+a2=a+6

Solution for 3a2+a2=a+6 equation:

3a^2+a2=a+6

We move all terms to the left:

3a^2+a2-(a+6)=0

We add all the numbers together, and all the variables

4a^2-(a+6)=0

We get rid of parentheses

4a^2-a-6=0

We add all the numbers together, and all the variables

4a^2-1a-6=0

a = 4; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·4·(-6)
Δ = 97
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{-(-1)-\sqrt{97}}{2*4}=\frac{1-\sqrt{97}}{8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{97}}{2*4}=\frac{1+\sqrt{97}}{8}
$