3y^2+6y=105

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Solution for 3y^2+6y=105 equation:



3y^2+6y=105
We move all terms to the left:
3y^2+6y-(105)=0
a = 3; b = 6; c = -105;
Δ = b2-4ac
Δ = 62-4·3·(-105)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-36}{2*3}=\frac{-42}{6} =-7 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+36}{2*3}=\frac{30}{6} =5 $

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