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10y^2+38y+24=0
a = 10; b = 38; c = +24;
Δ = b2-4ac
Δ = 382-4·10·24
Δ = 484
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{484}=22$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-22}{2*10}=\frac{-60}{20} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+22}{2*10}=\frac{-16}{20} =-4/5 $
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