A(x)=(1-x)(6x-8)+16-9x

Solution for A(x)=(1-x)(6x-8)+16-9x equation:

(A)=(1-A)(6A-8)+16-9A

We move all terms to the left:

(A)-((1-A)(6A-8)+16-9A)=0

We add all the numbers together, and all the variables

A-((-1A+1)(6A-8)+16-9A)=0

We multiply parentheses ..

-((-6A^2+8A+6A-8)+16-9A)+A=0


We calculate terms in parentheses: -((-6A^2+8A+6A-8)+16-9A), so:

(-6A^2+8A+6A-8)+16-9A

determiningTheFunctionDomain
(-6A^2+8A+6A-8)-9A+16

We get rid of parentheses

-6A^2+8A+6A-9A-8+16

We add all the numbers together, and all the variables

-6A^2+5A+8

Back to the equation:

-(-6A^2+5A+8)


We get rid of parentheses

6A^2-5A+A-8=0

We add all the numbers together, and all the variables

6A^2-4A-8=0

a = 6; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·6·(-8)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{13}}{2*6}=\frac{4-4\sqrt{13}}{12}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{13}}{2*6}=\frac{4+4\sqrt{13}}{12}
$