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

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



(3x+1)(2x-2)=0
We multiply parentheses ..
(+6x^2-6x+2x-2)=0
We get rid of parentheses
6x^2-6x+2x-2=0
We add all the numbers together, and all the variables
6x^2-4x-2=0
a = 6; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·6·(-2)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*6}=\frac{-4}{12} =-1/3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*6}=\frac{12}{12} =1 $

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