16x^2-4(x+1)(x+1)=0

Solution for 16x^2-4(x+1)(x+1)=0 equation:

16x^2-4(x+1)(x+1)=0

We multiply parentheses ..

16x^2-4(+x^2+x+x+1)=0

We multiply parentheses

16x^2-4x^2-4x-4x-4=0

We add all the numbers together, and all the variables

12x^2-8x-4=0

a = 12; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·12·(-4)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*12}=\frac{-8}{24}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*12}=\frac{24}{24}
=1
$