2n^2+3n-3=0

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



2n^2+3n-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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


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

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