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21y^2+23y-20=0
a = 21; b = 23; c = -20;
Δ = b2-4ac
Δ = 232-4·21·(-20)
Δ = 2209
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{2209}=47$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-47}{2*21}=\frac{-70}{42} =-1+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+47}{2*21}=\frac{24}{42} =4/7 $
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