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10y^2+60y+50=0
a = 10; b = 60; c = +50;
Δ = b2-4ac
Δ = 602-4·10·50
Δ = 1600
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{1600}=40$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-40}{2*10}=\frac{-100}{20} =-5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+40}{2*10}=\frac{-20}{20} =-1 $
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