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

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



(3y-1)(y+2)=0
We multiply parentheses ..
(+3y^2+6y-1y-2)=0
We get rid of parentheses
3y^2+6y-1y-2=0
We add all the numbers together, and all the variables
3y^2+5y-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:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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

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