3n^2+3n-4=0

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



3n^2+3n-4=0

a = 3; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·3·(-4)
Δ = 57
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{57}}{2*3}=\frac{-3-\sqrt{57}}{6}
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{57}}{2*3}=\frac{-3+\sqrt{57}}{6}
$

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