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3y^2-12y-7=0
a = 3; b = -12; c = -7;
Δ = b2-4ac
Δ = -122-4·3·(-7)
Δ = 228
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{57}}{2*3}=\frac{12-2\sqrt{57}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{57}}{2*3}=\frac{12+2\sqrt{57}}{6} $
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