(6x+1)(1-3x)=0

Solution for (6x+1)(1-3x)=0 equation:

(6x+1)(1-3x)=0

We add all the numbers together, and all the variables

(6x+1)(-3x+1)=0

We multiply parentheses ..

(-18x^2+6x-3x+1)=0

We get rid of parentheses

-18x^2+6x-3x+1=0

We add all the numbers together, and all the variables

-18x^2+3x+1=0

a = -18; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·(-18)·1
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*-18}=\frac{-12}{-36}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*-18}=\frac{6}{-36}
=-1/6
$