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20y^2+21y+4=0
a = 20; b = 21; c = +4;
Δ = b2-4ac
Δ = 212-4·20·4
Δ = 121
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{121}=11$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*20}=\frac{-32}{40} =-4/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*20}=\frac{-10}{40} =-1/4 $
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