(2a+4)(2+4)=(2a+4)(2a+4)

Solution for (2a+4)(2+4)=(2a+4)(2a+4) equation:

(2a+4)(2+4)=(2a+4)(2a+4)

We move all terms to the left:

(2a+4)(2+4)-((2a+4)(2a+4))=0

We add all the numbers together, and all the variables

(2a+4)6-((2a+4)(2a+4))=0

We multiply parentheses

12a-((2a+4)(2a+4))+24=0

We multiply parentheses ..

-((+4a^2+8a+8a+16))+12a+24=0


We calculate terms in parentheses: -((+4a^2+8a+8a+16)), so:

(+4a^2+8a+8a+16)

We get rid of parentheses

4a^2+8a+8a+16

We add all the numbers together, and all the variables

4a^2+16a+16

Back to the equation:

-(4a^2+16a+16)


We add all the numbers together, and all the variables

12a-(4a^2+16a+16)+24=0

We get rid of parentheses

-4a^2+12a-16a-16+24=0

We add all the numbers together, and all the variables

-4a^2-4a+8=0

a = -4; b = -4; c = +8;
Δ = b2-4ac
Δ = -42-4·(-4)·8
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*-4}=\frac{-8}{-8}
=1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*-4}=\frac{16}{-8}
=-2
$