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

Solution for 4a^2+16=2a^2+4a+8a+16 equation:

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

We move all terms to the left:

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

We get rid of parentheses

4a^2-2a^2-4a-8a-16+16=0

We add all the numbers together, and all the variables

2a^2-12a=0

a = 2; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·2·0
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{-(-12)-12}{2*2}=\frac{0}{4}
=0
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*2}=\frac{24}{4}
=6
$