-3x(3x+2)+1=-41

Solution for -3x(3x+2)+1=-41 equation:

-3x(3x+2)+1=-41

We move all terms to the left:

-3x(3x+2)+1-(-41)=0

We add all the numbers together, and all the variables

-3x(3x+2)+42=0

We multiply parentheses

-9x^2-6x+42=0

a = -9; b = -6; c = +42;
Δ = b2-4ac
Δ = -62-4·(-9)·42
Δ = 1548
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:

$\sqrt{\Delta}=\sqrt{1548}=\sqrt{36*43}=\sqrt{36}*\sqrt{43}=6\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{43}}{2*-9}=\frac{6-6\sqrt{43}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{43}}{2*-9}=\frac{6+6\sqrt{43}}{-18}
$