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2y^2+9y-26=0
a = 2; b = 9; c = -26;
Δ = b2-4ac
Δ = 92-4·2·(-26)
Δ = 289
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{289}=17$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-17}{2*2}=\frac{-26}{4} =-6+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+17}{2*2}=\frac{8}{4} =2 $
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