a(a+3)=3(a+2)

Solution for a(a+3)=3(a+2) equation:

a(a+3)=3(a+2)

We move all terms to the left:

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

We multiply parentheses

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


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

3(a+2)

We multiply parentheses

3a+6

Back to the equation:

-(3a+6)


We get rid of parentheses

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

We add all the numbers together, and all the variables

a^2-6=0

a = 1; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·1·(-6)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*1}=\frac{0-2\sqrt{6}}{2}
=-\frac{2\sqrt{6}}{2}
=-\sqrt{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*1}=\frac{0+2\sqrt{6}}{2}
=\frac{2\sqrt{6}}{2}
=\sqrt{6}
$