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10y^2+49y-9=0
a = 10; b = 49; c = -9;
Δ = b2-4ac
Δ = 492-4·10·(-9)
Δ = 2761
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-\sqrt{2761}}{2*10}=\frac{-49-\sqrt{2761}}{20} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+\sqrt{2761}}{2*10}=\frac{-49+\sqrt{2761}}{20} $
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