4x2+4x+1=16

Solution for 4x2+4x+1=16 equation:

4x^2+4x+1=16

We move all terms to the left:

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

We add all the numbers together, and all the variables

4x^2+4x-15=0

a = 4; b = 4; c = -15;
Δ = b2-4ac
Δ = 42-4·4·(-15)
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{-(4)-16}{2*4}=\frac{-20}{8}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*4}=\frac{12}{8}
=1+1/2
$