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4y^2+10y-104=0
a = 4; b = 10; c = -104;
Δ = b2-4ac
Δ = 102-4·4·(-104)
Δ = 1764
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{1764}=42$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-42}{2*4}=\frac{-52}{8} =-6+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+42}{2*4}=\frac{32}{8} =4 $
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