16x^2+8x+1=4

Solution for 16x^2+8x+1=4 equation:

16x^2+8x+1=4

We move all terms to the left:

16x^2+8x+1-(4)=0

We add all the numbers together, and all the variables

16x^2+8x-3=0

a = 16; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·16·(-3)
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*16}=\frac{-24}{32}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*16}=\frac{8}{32}
=1/4
$