3y^2+3y-5=0

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Solution for 3y^2+3y-5=0 equation: 



3y^2+3y-5=0

a = 3; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·3·(-5)
Δ = 69
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{69}}{2*3}=\frac{-3-\sqrt{69}}{6}
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{69}}{2*3}=\frac{-3+\sqrt{69}}{6}
$

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