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2u^2+23u+11=0
a = 2; b = 23; c = +11;
Δ = b2-4ac
Δ = 232-4·2·11
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-21}{2*2}=\frac{-44}{4} =-11 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+21}{2*2}=\frac{-2}{4} =-1/2 $
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