3x^2-2802x=1

Solution for 3x^2-2802x=1 equation:

3x^2-2802x=1

We move all terms to the left:

3x^2-2802x-(1)=0

a = 3; b = -2802; c = -1;
Δ = b2-4ac
Δ = -28022-4·3·(-1)
Δ = 7851216
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:

$\sqrt{\Delta}=\sqrt{7851216}=\sqrt{16*490701}=\sqrt{16}*\sqrt{490701}=4\sqrt{490701}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2802)-4\sqrt{490701}}{2*3}=\frac{2802-4\sqrt{490701}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2802)+4\sqrt{490701}}{2*3}=\frac{2802+4\sqrt{490701}}{6}
$