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2x^2+19x+5=0
a = 2; b = 19; c = +5;
Δ = b2-4ac
Δ = 192-4·2·5
Δ = 321
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{321}}{2*2}=\frac{-19-\sqrt{321}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{321}}{2*2}=\frac{-19+\sqrt{321}}{4} $
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