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

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



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

a = -2; b = 3; c = +3;
Δ = b2-4ac
Δ = 32-4·(-2)·3
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{33}}{2*-2}=\frac{-3-\sqrt{33}}{-4}
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*-2}=\frac{-3+\sqrt{33}}{-4}
$

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