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5y^2+25y+20=0
a = 5; b = 25; c = +20;
Δ = b2-4ac
Δ = 252-4·5·20
Δ = 225
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{225}=15$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-15}{2*5}=\frac{-40}{10} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+15}{2*5}=\frac{-10}{10} =-1 $
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