(3a+2)(a-3)=(a-3)(a+4)

Solution for (3a+2)(a-3)=(a-3)(a+4) equation:

(3a+2)(a-3)=(a-3)(a+4)

We move all terms to the left:

(3a+2)(a-3)-((a-3)(a+4))=0

We multiply parentheses ..

(+3a^2-9a+2a-6)-((a-3)(a+4))=0


We calculate terms in parentheses: -((a-3)(a+4)), so:

(a-3)(a+4)

We multiply parentheses ..

(+a^2+4a-3a-12)

We get rid of parentheses

a^2+4a-3a-12

We add all the numbers together, and all the variables

a^2+a-12

Back to the equation:

-(a^2+a-12)


We get rid of parentheses

3a^2-a^2-9a+2a-a-6+12=0

We add all the numbers together, and all the variables

2a^2-8a+6=0

a = 2; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*2}=\frac{4}{4}
=1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*2}=\frac{12}{4}
=3
$