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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((-6A^2+10A+9A-15))=0


We calculate terms in parentheses: -((-6A^2+10A+9A-15)), so:

(-6A^2+10A+9A-15)

We get rid of parentheses

-6A^2+10A+9A-15

We add all the numbers together, and all the variables

-6A^2+19A-15

Back to the equation:

-(-6A^2+19A-15)


We get rid of parentheses

6A^2-19A+15=0

a = 6; b = -19; c = +15;
Δ = b2-4ac
Δ = -192-4·6·15
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*6}=\frac{18}{12}
=1+1/2
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*6}=\frac{20}{12}
=1+2/3
$