+3y^2+y-36=0

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Solution for +3y^2+y-36=0 equation: 



+3y^2+y-36=0

a = 3; b = 1; c = -36;
Δ = b2-4ac
Δ = 12-4·3·(-36)
Δ = 433
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{-(1)-\sqrt{433}}{2*3}=\frac{-1-\sqrt{433}}{6}
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{433}}{2*3}=\frac{-1+\sqrt{433}}{6}
$

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