A=(x+3)2+(5x+2)(x-3)

Solution for A=(x+3)2+(5x+2)(x-3) equation:

=(A+3)2+(5A+2)(A-3)

We move all terms to the left:

-((A+3)2+(5A+2)(A-3))=0

We multiply parentheses ..

-((A+3)2+(+5A^2-15A+2A-6))=0


We calculate terms in parentheses: -((A+3)2+(+5A^2-15A+2A-6)), so:

(A+3)2+(+5A^2-15A+2A-6)

determiningTheFunctionDomain
(+5A^2-15A+2A-6)+(A+3)2

We multiply parentheses

(+5A^2-15A+2A-6)+2A+6

We get rid of parentheses

5A^2-15A+2A+2A-6+6

We add all the numbers together, and all the variables

5A^2-11A

Back to the equation:

-(5A^2-11A)


We get rid of parentheses

-5A^2+11A=0

a = -5; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·(-5)·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*-5}=\frac{-22}{-10}
=2+1/5
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*-5}=\frac{0}{-10}
=0
$