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3y^2-10y-8=0
a = 3; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·3·(-8)
Δ = 196
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{196}=14$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*3}=\frac{-4}{6} =-2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*3}=\frac{24}{6} =4 $
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