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

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

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

We move all terms to the left:

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

We multiply parentheses

3A^2+(-3)(A-3)-6A-(4(A+2))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -(4(A+2)), so:

4(A+2)

We multiply parentheses

4A+8

Back to the equation:

-(4A+8)


We add all the numbers together, and all the variables

3A^2-6A+(-3A+9)-(4A+8)=0

We get rid of parentheses

3A^2-6A-3A-4A+9-8=0

We add all the numbers together, and all the variables

3A^2-13A+1=0

a = 3; b = -13; c = +1;
Δ = b2-4ac
Δ = -132-4·3·1
Δ = 157
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{157}}{2*3}=\frac{13-\sqrt{157}}{6}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{157}}{2*3}=\frac{13+\sqrt{157}}{6}
$