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3y^2-20y-63=0
a = 3; b = -20; c = -63;
Δ = b2-4ac
Δ = -202-4·3·(-63)
Δ = 1156
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{1156}=34$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-34}{2*3}=\frac{-14}{6} =-2+1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+34}{2*3}=\frac{54}{6} =9 $
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