(2a2-4a)+1=1

Solution for (2a2-4a)+1=1 equation:

(2a^2-4a)+1=1

We move all terms to the left:

(2a^2-4a)+1-(1)=0

We add all the numbers together, and all the variables

(2a^2-4a)=0

We get rid of parentheses

2a^2-4a=0

a = 2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*2}=\frac{0}{4}
=0
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*2}=\frac{8}{4}
=2
$