3a^+a^2=a+6

Solution for 3a^+a^2=a+6 equation:

3a^+a^2=a+6

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

a^2+3a-a-6=0

We add all the numbers together, and all the variables

a^2+2a-6=0

a = 1; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·1·(-6)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{7}}{2*1}=\frac{-2-2\sqrt{7}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{7}}{2*1}=\frac{-2+2\sqrt{7}}{2}
$