a2+6a+9=(a+3)2

Solution for a2+6a+9=(a+3)2 equation:

a2+6a+9=(a+3)2

We move all terms to the left:

a2+6a+9-((a+3)2)=0

We add all the numbers together, and all the variables

a^2+6a-((a+3)2)+9=0


We calculate terms in parentheses: -((a+3)2), so:

(a+3)2

We multiply parentheses

2a+6

Back to the equation:

-(2a+6)


We get rid of parentheses

a^2+6a-2a-6+9=0

We add all the numbers together, and all the variables

a^2+4a+3=0

a = 1; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*1}=\frac{-6}{2}
=-3
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*1}=\frac{-2}{2}
=-1
$