12x^2-28x=1

Solution for 12x^2-28x=1 equation:

12x^2-28x=1

We move all terms to the left:

12x^2-28x-(1)=0

a = 12; b = -28; c = -1;
Δ = b2-4ac
Δ = -282-4·12·(-1)
Δ = 832
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-8\sqrt{13}}{2*12}=\frac{28-8\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+8\sqrt{13}}{2*12}=\frac{28+8\sqrt{13}}{24}
$