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2y^2+11y+5=0
a = 2; b = 11; c = +5;
Δ = b2-4ac
Δ = 112-4·2·5
Δ = 81
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{81}=9$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-9}{2*2}=\frac{-20}{4} =-5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+9}{2*2}=\frac{-2}{4} =-1/2 $
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