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x^2+19x-68=0
a = 1; b = 19; c = -68;
Δ = b2-4ac
Δ = 192-4·1·(-68)
Δ = 633
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{633}}{2*1}=\frac{-19-\sqrt{633}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{633}}{2*1}=\frac{-19+\sqrt{633}}{2} $
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