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2x^2+19x+39=0
a = 2; b = 19; c = +39;
Δ = b2-4ac
Δ = 192-4·2·39
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-7}{2*2}=\frac{-26}{4} =-6+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+7}{2*2}=\frac{-12}{4} =-3 $
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