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2y^2+27y+13=0
a = 2; b = 27; c = +13;
Δ = b2-4ac
Δ = 272-4·2·13
Δ = 625
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{625}=25$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-25}{2*2}=\frac{-52}{4} =-13 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+25}{2*2}=\frac{-2}{4} =-1/2 $
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