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2y^2+23y+65=0
a = 2; b = 23; c = +65;
Δ = b2-4ac
Δ = 232-4·2·65
Δ = 9
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{9}=3$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-3}{2*2}=\frac{-26}{4} =-6+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+3}{2*2}=\frac{-20}{4} =-5 $
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