3x^2+27x=1

Solution for 3x^2+27x=1 equation:

3x^2+27x=1

We move all terms to the left:

3x^2+27x-(1)=0

a = 3; b = 27; c = -1;
Δ = b2-4ac
Δ = 272-4·3·(-1)
Δ = 741
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-\sqrt{741}}{2*3}=\frac{-27-\sqrt{741}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{741}}{2*3}=\frac{-27+\sqrt{741}}{6}
$