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

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



(-3x+2)(1x+1)=0
We add all the numbers together, and all the variables
(-3x+2)(x+1)=0
We multiply parentheses ..
(-3x^2-3x+2x+2)=0
We get rid of parentheses
-3x^2-3x+2x+2=0
We add all the numbers together, and all the variables
-3x^2-1x+2=0
a = -3; b = -1; c = +2;
Δ = b2-4ac
Δ = -12-4·(-3)·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-3}=\frac{-4}{-6} =2/3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-3}=\frac{6}{-6} =-1 $

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