3y^2+18=102

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Solution for 3y^2+18=102 equation:



3y^2+18=102
We move all terms to the left:
3y^2+18-(102)=0
We add all the numbers together, and all the variables
3y^2-84=0
a = 3; b = 0; c = -84;
Δ = b2-4ac
Δ = 02-4·3·(-84)
Δ = 1008
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{7}}{2*3}=\frac{0-12\sqrt{7}}{6} =-\frac{12\sqrt{7}}{6} =-2\sqrt{7} $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{7}}{2*3}=\frac{0+12\sqrt{7}}{6} =\frac{12\sqrt{7}}{6} =2\sqrt{7} $

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