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8u^2+19u+6=0
a = 8; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·8·6
Δ = 169
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{169}=13$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*8}=\frac{-32}{16} =-2 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*8}=\frac{-6}{16} =-3/8 $
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