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5y^2+40y-45=0
a = 5; b = 40; c = -45;
Δ = b2-4ac
Δ = 402-4·5·(-45)
Δ = 2500
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{2500}=50$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-50}{2*5}=\frac{-90}{10} =-9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+50}{2*5}=\frac{10}{10} =1 $
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