-3y^2+8y-4=0

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Solution for -3y^2+8y-4=0 equation:



-3y^2+8y-4=0
a = -3; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·(-3)·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*-3}=\frac{-12}{-6} =+2 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*-3}=\frac{-4}{-6} =2/3 $

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