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3y^2-4y-288=0
a = 3; b = -4; c = -288;
Δ = b2-4ac
Δ = -42-4·3·(-288)
Δ = 3472
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3472}=\sqrt{16*217}=\sqrt{16}*\sqrt{217}=4\sqrt{217}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{217}}{2*3}=\frac{4-4\sqrt{217}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{217}}{2*3}=\frac{4+4\sqrt{217}}{6} $
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