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

Solution for (1-3x)(2+x)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2-5x+2=0

a = -3; b = -5; c = +2;
Δ = b2-4ac
Δ = -52-4·(-3)·2
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*-3}=\frac{-2}{-6}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*-3}=\frac{12}{-6}
=-2
$