a2+6a+6=40

Solution for a2+6a+6=40 equation:

a2+6a+6=40

We move all terms to the left:

a2+6a+6-(40)=0

We add all the numbers together, and all the variables

a^2+6a-34=0

a = 1; b = 6; c = -34;
Δ = b2-4ac
Δ = 62-4·1·(-34)
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{43}}{2*1}=\frac{-6-2\sqrt{43}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{43}}{2*1}=\frac{-6+2\sqrt{43}}{2}
$