3y2+8=72

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Solution for 3y2+8=72 equation:



3y^2+8=72
We move all terms to the left:
3y^2+8-(72)=0
We add all the numbers together, and all the variables
3y^2-64=0
a = 3; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·3·(-64)
Δ = 768
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:
$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*3}=\frac{0-16\sqrt{3}}{6} =-\frac{16\sqrt{3}}{6} =-\frac{8\sqrt{3}}{3} $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*3}=\frac{0+16\sqrt{3}}{6} =\frac{16\sqrt{3}}{6} =\frac{8\sqrt{3}}{3} $

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