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2x^2+21x+19=0
a = 2; b = 21; c = +19;
Δ = b2-4ac
Δ = 212-4·2·19
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-17}{2*2}=\frac{-38}{4} =-9+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+17}{2*2}=\frac{-4}{4} =-1 $
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