64x^2+12x=1

Solution for 64x^2+12x=1 equation:

64x^2+12x=1

We move all terms to the left:

64x^2+12x-(1)=0

a = 64; b = 12; c = -1;
Δ = b2-4ac
Δ = 122-4·64·(-1)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-20}{2*64}=\frac{-32}{128}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20}{2*64}=\frac{8}{128}
=1/16
$