A=3(2-4K)+k2

Solution for A=3(2-4K)+k2 equation:

=3(2-4)+A2

We move all terms to the left:

-(3(2-4)+A2)=0

We add all the numbers together, and all the variables

-(3(-2)+A2)=0


We calculate terms in parentheses: -(3(-2)+A2), so:

3(-2)+A2

determiningTheFunctionDomain
A2+3(-2)

We add all the numbers together, and all the variables

A^2-6

Back to the equation:

-(A^2-6)


We get rid of parentheses

-A^2+6=0

We add all the numbers together, and all the variables

-1A^2+6=0

a = -1; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-1)·6
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-1}=\frac{0-2\sqrt{6}}{-2}
=-\frac{2\sqrt{6}}{-2}
=-\frac{\sqrt{6}}{-1}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-1}=\frac{0+2\sqrt{6}}{-2}
=\frac{2\sqrt{6}}{-2}
=\frac{\sqrt{6}}{-1}
$