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14x^2+19x-8=0
a = 14; b = 19; c = -8;
Δ = b2-4ac
Δ = 192-4·14·(-8)
Δ = 809
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{809}}{2*14}=\frac{-19-\sqrt{809}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{809}}{2*14}=\frac{-19+\sqrt{809}}{28} $
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